Tài liệu Thủ Thuật Tính Tích Phân Cực Hay

Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng MËT SÈ Kž THUŠT "PH…N TCH " M€ TA TH×ÍNG GP KHI I TœM NGUY–N H€M HOC TNH ZTCH PH…N. Th½ dö 1 : T¼m nguy¶n h m A1 = (x + 1)dx(x + 2) Ta gi£ sû r¬ng : (x + 1)1(x + 2) = x +α 1 + x +β 2 Ta i t¼m 2 h» sè α , β theo ba c¡ch nh÷ sau : x+1 1 1 C¡ch 1 : α = x→−1 lim = lim = =1 (x + 1) (x + 2) x→−1 x + 2 −1 + 2 x+2 1 1 = lim = = −1 x→−2 (x + 1) (x + 2) x→−2 x + 1 −2 + 1 β = lim C¡ch 2 : Cho x = 0 ta câ : (0 + 1)1(0 + 2) = 0 +α 1 + 0 +β 2 ⇔ 12 = α + 12 β x=1 ta câ : (1 + 1)1(1 + 2) = 1 +α 1 + 1 +β 2 ⇔ 16 = 12 α + 31 β   α + 1β = 1 Do â m  ta suy ra 2 h» sè α , β b¬ng c¡ch i gi£i h» :  C¡ch 3 : Ta gi£ sû r¬ng : (x + 1)1(x + 2) = x +α 1 + x +β 2 ⇔ ( 2 2 1 1 ⇔ 1 α+ β = 2 3 6 α=1 β = −1 1 α (x + 2) + β (x + 1) x (α + β) + 2α + β = = (x + 1) (x + 2) (x + 1) (x + 2) ( (x + 1) (x + (2) C¥n b¬ng c¡c h» sè 2 v¸ m  ta câ h» : α+β =0 ⇔ α=1 β = −1 2α + β = 1 Do â m  ta suy ra : (x + 1)1(x + 2) = x +1 1 − x +1 2 Z Z Z   dx dx dx x + 1 Vªy : (x + 1) (x + 2) = x + 1 − x + 2 = ln |x + 1| − ln |x + 2| + c = ln  x + 2  + c Z 2 : T¼m nguy¶n h m A2 = x (x −x2)+(x dx + 5) 2 α β χ Ta gi£ sû r¬ng : x (x −x2)+(x = + + + 5) x x−2 x+5 Th½ dö 2 C¡ch 1 : x (x + 2) x+2 0+2 1 = lim = =− x→0 x (x − 2) (x + 5) x→0 (x − 2) (x + 5) (0 − 2) (0 + 5) 5 (x − 2) (x + 2) x+2 2+2 2 β = lim = lim = = x→2 x (x − 2) (x + 5) x→2 x (x + 5) 2 (2 + 5) 7 (x + 5) (x + 2) x+2 −5 + 2 3 χ = lim = lim = =− x→−5 x (x − 2) (x + 5) x→−5 x (x − 2) −5 (−5 − 2) 35 α = lim C¡ch 2 : +2 α β χ 1 1 1 Cho x = −1 ta câ : −1 (−1−1 = + + ⇔ −α − β + χ = − 2) (−1 + 5) −1 −1 − 2 −1 + 5 3 4 12 x=1 ta câ : 1 (1 −12)+(12 + 5) = α1 + 1 −β 2 + 1 +χ 5 ⇔ α − β + 61 χ = − 12 1 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng ta câ : 3 (3 −32)+(32 + 5) = α3 + 3 −β 2 + 3 +χ 5 ⇔ 13 α + β + 81 χ = 245 º t¼m ÷ñc c¡c h» sèα , β , χ ta gi£i h» ph÷ìng tr¼nh  : x=3 1 1 1  −α − β + χ =   3 4 12      1 1 α−β+ χ=− 6 2 1 1 5 α+β+ χ= 3 8 24 ⇔ 1  α=−   5  2 7 3 χ=− 35 β=     C¡ch 3 : α β χ 2 = + + Gi£ sû r¬ng : x (x −x2)+(x + 5) x x−2 x+5 α (x − 2) (x + 5) + βx (x + 5) + χx (x − 2) x+2 = x (x − 2) (x + 5) x (x − 2) (x + 5) 2 x (α + β + χ) + x (3α + 5β − 2χ) − 10α x+2 = ⇔ x (x − 2) (x + 5) x (x − 2) (x + 5)  1   α=−    α + β + χ = 0 5   2 C¥n b¬ng c¡c h» sè 2 v¸ m  ta câ h» :  3α + 5β − 2χ = 1 ⇔  β = 7  −10α = 2    χ=−3 35 x+2 1 2 3 Do â : x (x − 2) (x + 5) = − 5x + 7 (x − 2) − 35 (x + 5) Z Z Z Z x+2 1 dx 2 dx 3 Vªy : x (x − 2) (x + 5) dx = − 5 x + 7 x − 2 − 35 xdx +5 2 3 1 ln |x + 5| + c = − ln |x| + ln |x − 2| − 5 Z 7 35 2 : T¼m nguy¶n h m A3 = (−3x2 − 2xx+ 5) (x + 1) dx Ta c¦n nhî ph÷ìng tr¼nh bªc 2 câ d¤ng : ax2 + bx + c câ 2 nghi»m x1, , x2 th¼ ÷ñc biºu di¹n d÷îi ⇔ Th½ dö 3 d¤ng : ax2 + bx + c = a (x − x1) (x − x2)   5 2 Do â ta vi¸t : −3x − 2x + 5 = −3 (x − 1) x + 3 x2 2 Ta gi£ sû r¬ng : (−3x2 − 2xx+ 5) (x + 1) = −   5 3 (x − 1) x + (x + 1) 3     = α + x−1 β 5 x+ 3 + χ x+1 x2 (x − 1) x2 1    = lim − =−        5 5 x→1 x→1 16 3 (x − 1) x + (x + 1) 3 x+ (x + 1) 3    3 5   x2 x + x2 25   3 β = lim − = lim − =−    5 3 (x − 1) (x + 1) 48 x→− 53 x→− 53 3 (x − 1) x + (x + 1) 3     α = lim −  χ = lim −  x→−1 C¡ch 2 : x2 (x + 1) x2    1 = lim − =      5 5  4 x→−1 3 (x − 1) x + (x + 1) 3 (x − 1) x + 3 3 2 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng 2 Cho x = 0 ta câ : (−3x2 − 2xx+ 5) (x + 1) = x −α 1 + º χ 3 + ⇔ −α + β + χ = 0 5 x+1 5 x+ 3 χ x2 α β 4 3 1 + x = 2 ta câ : = + ⇔− =α+ β+ χ 2 5 (−3x − 2x + 5) (x + 1) x−1 x+1 33 11 3 x+ 3 α β χ 9 1 3 1 x2 = + + ⇔− = α+ β+ χ x = 3 ta câ : 2 5 (−3x − 2x + 5) (x + 1) x−1 x+1 112 2 14 4 x+ 3   3 1   −α + β + χ = 0 α=−     5 16   25 tø â m  ta câ h» :  α + 3 β + 1 χ = − 4 ⇔ β=−  11 3 33 48      1α + 3 β + 1χ = − 9  χ= 1 2 14 4 112 4 C¡ch 3 : 2 Gi£ sû r¬ng : (−3x2 − 2xx+ 5) (x + 1) = − β x2 Do â m  ta2 câ : x =− 2 (−3x − 2x + 5) (x + 1) 3 (x − 1) x +  5 = α + x−1 β + χ x+1 5 3    3 5 5 α x+ (x + 1) + β (x − 1) (x + 1) + χ (x − 1) x + x2 3 3 = ⇔−     5 5 3 (x − 1) x + (x + 1) (x − 1) x + (x + 1) 3 3 8  2 5 5 ⇔ x2 = x2 (α + β + χ) + x α+ χ + α−β− χ 3 3 3 3   1  α+β+χ=1 α=−     16  8  2 25 α + χ = 0 ⇔ C¥n b¬ng c¡c h» sè 2 v¸ ta câ h» :  3 3 β=−  48   5 5  α−β− χ=0   χ= 1 3 3 4  (x + 1) x+ x2 1 25 1 =− −   + 5 5 16 (x − 1) 4 (x + 1) 3 (x − 1) x + (x + 1) 48 x + 3 Z 3 Z Z Z x2 1 dx 25 dx 1 dx + Suy ra : (−3x2 − 2x + 5) (x + 1) dx = − 16 x − 1 − 48 5 4 x+4 x+ 3   1 25  5 1 = − ln |x − 1| − ln x +  + ln |x + 4| + c 16 48 3 4 Z x−1 : T¼m nguy¶n h m A4 = (x2 + 4x + 5) (x2 − 4) dx Ta c¦n chó þ r¬ng ph÷ìng tr¼nh : ax2 + bx + c = 0 vîi ∆ = b2 − 4ac < 0 th¼ ta vi¸t :  Th½ dö 4 ax2 + bx + c = a (x − x1 ) (x − x2 ) trong â : x1 = α + βi, x2 = α − βi, i2 = −1 Do â m  ta câ : x2 + 4x + 5 = (x + 2 + i) (x + 2 − i) Ta gi£ sû r¬ng : x−1 x−1 α β χ δ = = + + + 2 2 + 4x + 5) (x − 4) (x + 2 + i) (x + 2 − i) (x − 4) x+2+i x+2−i x−2 x+2 (x − 1) (x + 2 + i) x−1 13 1 α = lim = lim =− − i 2 2 x→−2−i (x + 2 + i) (x + 2 − i) (x − 4) x→−2−i (x + 2 − i) (x − 4) 34 34 (x2 3 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng (x − 1) (x + 2 − i) x−1 13 1 = lim = − + i x→−2+i (x + 2 + i) (x + 2 − i) (x2 − 4) x→−2+i (x + 2 + i) (x2 − 4) 34 34 (x − 1) (x − 2) x−1 1 χ = lim 2 = lim 2 = x→2 (x + 4x + 5) (x − 2) (x + 2) x→2 (x + 4x + 5) (x + 2) 68 x−1 3 (x − 1) (x + 2) = lim = δ = lim 2 2 x→−2 (x + 4x + 5) (x − 2) x→−2 (x + 4x + 5) (x − 2) (x + 2) 4 13 13 1 3 1 1 − − i − + i x−1 Do â m  ta câ : (x2 + 4x + 5) (x2 − 4) = x34+ 2 +34i + x34+ 2 −34i + x 68 + 4 −2 x+2 −13x − 27 1 3 = + + 2 17 (x + 4x + 5) 68 (x − 2) 4 (x + 2) β= lim C¡ch 2 : +β χ δ + + Gi£ sû r¬ng : (x2 + 4xx+−5)1 (x2 − 4) = x2αx + 4x + 5 x − 2 x + 2 Cho x = 0 ta câ : 51 β − 21 χ + 12 δ = 201 1 1 1 x = 1 ta câ : α+ β−χ+ δ =0 10 10 3 3 1 1 1 x = 3 ta câ : α+ β+χ+ δ = 26 26 5 65 4 1 1 1 1 x = 4 ta câ : α+ β+ χ+ δ = 37 37 2 6 148 º tø â m  ta câ ÷ñc h» :   1 1 1 1  β− χ+ δ =   5 2 2 20     1 α + 1 β − χ + 1δ = 0 10 10 3 ⇔ 1 1 1 3   α+ β+χ+ δ =   26 26 5 65   1 1 1 1  4 37 1 2  δ =− β+χ+   5 10     1 α − 1 β − 2χ = − 1 10 30 3 30 3 27 6 3   α− β+ χ=−   26 650 5 650   22 2 11  4 ⇔  13  α=−   17     β = − 27 β+ χ+ δ = α− β+ χ=− 2 6 148 37 555 3 1110 x−1 −13x − 27β χ 3δ = + + 2 2 2 (x + 4x + 5) (x − 4) 17 (x + 4x + 5) 68 (x − 2) 4 (x + 2) α+ 37 17 1   χ=   68   3  δ= Do â : C¡ch 3 : +β χ δ + + Gi£ sû : (x2 + 4xx+−5)1 (x2 − 4) = x2αx + 4x + 5 x − 2 x + 2 4 ⇔ x − 1 = (αx + β) x2 − 4 + χ x2 + 4x + 5 (x + 2) + δ x2 + 4x + 5 (x − 2)


⇔ x − 1 = x3 (α + χ + δ) + x2(β + 6χ + 2δ) + x (−4α + 13χ  − 3δ) + (−4β + 10χ − 10δ)    δ = −α − χ  α+χ+δ =0 C¥n b¬ng c¡c h» sè 2 v¸ ta câ h» :    β + 6χ + 2 (−α − χ) = 0  β + 6χ + 2δ = 0 ⇔   −4α + 13χ − 3δ = 1 −4α + 13χ − 3 (−α − χ) = 1       4β − 10χ + 10δ = 1 4β − 10χ + 10 (−α − χ) = 1  13   α=−   17  δ = −α − χ    27     ⇔ −2α + β + 4χ = 0  −α + 16χ = 1    −10α + 4β − 20χ = 1 ⇔        β=− 17 1 χ= 68 3 δ= 4 4 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng −13x − 27 1 3 Do â : (x2 + 4xx+−5)1 (x2 − 4) = 17 (x + + 2 + 4x + 5) 68 (x − 2) 4 (x + 2) Z 2x + 54 Z Z x−1 13 1 dx 3 13 Vªy : (x2 + 4x + 5) (x2 − 4) dx = − 34 x2 + 4x + 5 dx + 68 x − 2 + 4 xdx +2 Z Z Z Z (2x + 4) 1 dx 1 dx 3 dx 13 dx − + + =− 2 2 34 x + 4x + 5
17 x + 4x + 5 68 x−2 4 x+2 Z Z Z Z 2 d x + 4x + 5 13 1 1 dx dx 3 dx =− + − + 2 2 34 x + 4x + 5 17 x−2 4 x+2 (x + 2) + 1 68   1 1 3 13 arctan (x + 2) + ln |x − 2| + ln |x + 2| + c = − ln x2 + 4x + 5 − 34 17 Z 68 4 x : T¼m nguy¶n h m A5 = x3 + 1 dx β χ √ + √ Ta gi£ sû r¬ng : x3 x+ 1 = (x + 1) (xx2 − x + 1) = x +α 1 + 1 3 1 3 x− − i x− + i 2 2 2 2 1 x =− α = lim 2 x→−1 x − x + 1 3 √ 1 x 3  √ = − β = lim√ i 6 6 1 3 x→ 21 + 23 i i (x + 1) x − + 2 2 √ x 1 3  √ = + χ = lim√ i 6 6 1 3 x→ 12 − 23 i (x + 1) x − − i 2 2 √ √ 3 1 3 1 − i + i x 1 Do â : x3 + 1 = − 3 (x + 1) + 6 6√ + 6 6√ = − 3 (x1+ 1) + 3 (x2x−+x1+ 1) 1 1 3 3 x− − i x− + i 2 2 2 2 Z Th½ dö 5 C¡ch 2 : Gi£ sû : x3 x+ 1 = x +α 1 + x2βx− +x +χ 1 Cho x = 0 ta câ : 0 = α + χ 1 1 x = 1 ta câ : = α + β + χ 2 2 2 1 2 1 x = 2 ta câ : = α + β + χ 9 3 3 3  1  α=−   3   α+χ=0    1 1 α+β+χ= ⇔ β= 2 2    1 2 1 2  α+ β+ χ=   χ= 3 3 3 9 x 1 x+1 = − + x3 + 1 3 (x + 1) 3 (x2 − x + 1) Do â m  ta câ h» :  Vªy : C¡ch 3 : Ta công gi£ sû r¬ng : x3 x+ 1 = x +α 1 + x2βx− +x +χ 1 1 3 1 3 ⇔ x = α x2 − x + 1 + (βx + χ) (x + 1)
5 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng    α+β =0 C¥n b¬ng c¡c h» sè m  ta câ h» :  −α + β + χ = 1  α+χ=0 Z Z Z ⇔  1  α=−   3  1 3 1 χ= 3 β=     1 2x + 2 + dx Vªy : x3 x+ 1 dx = − 13 xdx 2 Z +1 6 Z x −x+1 Z 1 dx 1 2x + 1 1 dx =− + dx + 2 2 3 x+1 6 x −x+1
6 x −x+1 Z Z Z 2 d x −x+1 dx 1 1 dx 1 + + =−   3 x+1 6 x2 − x + 1 6 1 2 3 x− + 2 4   1 1  2 1 2x − 1  = − ln |x + 1| + ln x − x + 1 + √ arctan √ +c 3 3 3 3 Z6 : T¼m nguy¶n h m A6 = x4dx+ 1


Gi£ sû r¬ng : x4 + 1 = x2 + px + 1 x2 − √px + 1 = x4 + 2 − p2 x + 1 çng nh§t 2 v¸ ta câ : 2 − p2 =√0 ⇔ p =
± 2 √
Do â ta vi¸t : x4 + 1 = x2 + 2x + 1 x2 − 2x + 1 1

= √ √ ⇒ 2 x + 2x + 1 x2 − 2x + 1 α β χ δ √ + √ + √ + √ √ √ √ √ = 2 2 2 2 2 2 2 2 + i x+ − i x− + i x− − i x+ 2 2 2 2 2 2 2 2 T¼m c¡c h» sè : α, β, χ, δ nh÷ sau :  Th½ dö 6 √ √ √ √ √ √ √ √   α= lim √ √  x→− 2 − 2 i 1   = 2 + 2i
 8 8   β= lim √ √ 2 2  x→− + i 1   = 2 − 2i
 8 8 2 √  √ √ 2 2 − i x2 − 2x + 1 x+ 2 2   2 2 2    χ= lim √ √ 2 2  x→ − i 2 2   δ= lim √  √  x→ 2 + 2 i Vªy : √ √  √ 2 2 x+ + i x2 − 2x + 1 2 2  √ 1   = − 2 + 2i
 8 8 1   = − 2 − 2i
 8 8 √  √ 2 2 x− − i x2 + 2x + 1 2 2  √ √  √ 2 2 2 2 x− + i x2 + 2x + 1  √ 2 √ 2 √ √  √ √  √ √  2 2 2 2 2 2 2 2 + i x+ − i + − i x+ + i 8 8 2 2 8 8 2 2 1 √ = + x4 + 1 x2 + 2x + 1 6 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng √  √ √   √ √  √ √  2 2 2 2 2 2 2 2 + i x− − i + − − i x− + i − 8 8 2 2 8 8 2 2 √ + 2 − 2x + 1 x √ √ 1 1 2 2 x+ x+ − 4√ 2 + 4√ 2 = 2 2 x + 2x + 1 x − 2x + 1 C¡ch 2 : Ta gi£ sû r¬ng : x4 1+ 1 = 2 αx√+ β + 2 χx√+ δ x + 2x + 1 x − 2x + 1 1 1 Cho x = −1 ta câ : 2 = − √ α + 1√ β − 1√ χ + 1√ δ 2− 2 2− 2 2+ 2 2+ 2 x = 0 ta câ : β + δ = 1 1 1 1 1 1 √ α+ √ β− √ χ+ √ δ x = 1 ta câ : = 2 2+ 2 2+ 2 2− 2 2− 2 √

√ 1 1 √ 2α + β + 2χ + δ x = 2 ta câ : = 5 5  β + δ=1     1 1 1 1 1   √ α+ √ β− √ χ+ √ δ= − 2 º tø â ta câ h» :  21− 2 21− 2 21+ 2 21+ 2 1 √ α+ √ β− √ χ+ √ δ=   2 2 + 2 2 + 2 2 − 2 2 − 2   √ √

 1 1  2α + β + 2χ + δ = 5  5 β+δ =1   √ √    2α − 2β + 2χ + 2δ = 0 √ ⇔ √
√
2 + 2  α+β+ 3+2 2 χ+ 3+2 2 δ =   2  √ √ 2α + β + 5 2χ + 5δ = 1   √ 2 β + δ = 1     α= √ √     4   2α − 2β + 2χ = − 2 1 √ ⇔ ⇔ β=δ= √
√
4 + 3 2   √2 a− 2+2 2 β+ 3+2 2 χ=−     2   √ √   χ=− 2 2α − 4β + 5 2χ = −4 4 √ √ 2 1 2 1 x+ − x+ Vªy ta câ ÷ñc : x4 1+ 1 = 2 4 √ 2 + 2 4√ 2 x + 2x + 1 x − 2x + 1  √ C¡ch 3 : Ta công gi£ sû r¬ng : x4 1+ 1 = αx + β χx + δ √ √ + 2 + 2x + 1
x − 2x + 1 √ √
⇔ 1 = (αx + β) x2 − 2x + 1 + (χx + δ) x2 + 2x + 1 √ √ √
√
⇔ 1 = (α + χ) x3 + − 2α + β + 2χ + δ x2 + α − 2β + χ + 2δ x + β + δ x2 C¥n b¬ng c¡c h» sè 2 v¸ m  ta câ ÷ñc h» sau:   √ 2   α=   4    α+χ=0 α+χ=0      −√2α + β + √2χ + δ = 0  −√2α + √2χ = −1 1 √ √ √ √ ⇔ ⇔ β=δ=    α − 2β + χ + 2δ = 0 α − 2 2β + χ = − 2 √2           χ=− 2 β+δ =1 δ =1−β 4 7 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng √ 1 1 2 2 − x+ x+ 1 Do â : x4 + 1 = 2 4 √ 2 + 2 4√ 2 x + 2x + 1 x − 2x + 1 1 1 1 1 Z Z Z − √ x+ √ x+ Vªy ta câ : x4dx+ 1 = 22 2√ 2 dx + 2 2 √2 2 dx x + 2x + 1 x − 2x + 1 1 1 √ Z Z Z √ x+ 1 1 2x + 2 dx 2 2 2 √ √ dx = √ dx + Ta câ : 2 √ 4 x + Z 2x + 1 √ 4 2
x2 Z + 2x + 1 x2 + 2x + 1 2 d x + 2x + 1 1 1 dx √ = √ +  2 4 4 2 x2 + 2x + 1 1 1 x+ √ + 2 2 √ √

1 1 = √ ln x2 + 2x + 1 + √ arctan 2x + 1 + c1 4 2 2 2 1 1 √ Z Z Z − √ x+ 1 −2x + 2 1 dx 2 2 2 √ √ dx = √ dx + L¤i câ : 2 √ 2 2 4 x − Z2x + 1 x − 2x + 1 √ 4 2
x −Z 2x + 1 2 d x − 2x + 1 dx 1 1 √ =− √ +  2 4 4 2 x2 + 2x + 1 1 1 x− √ + 2 2 √ √

1 1 = − √ ln x2 − 2x + 1 + √ arctan 2x − 1 + c2 4 2 2  2 √ 2  Z √ √

dx 1 x + 2x + 1 1 1 Vªy : x4 + 1 = √ ln 2 √ + √ arctan 2x + 1 + √ arctan 2x − 1 + c 4 2 x − 2x 2 2 2 2 Z +1 dx : T¼m nguy¶n h m A7 = x8 + 1


Gi£ sû r¬ng : x8 + 1 = x4 + px2 + 1 x4 − px√2 + 1 = x8 + 2 − p2 x4 + 1 çng nh§t 2 v¸ ta √÷ñc : 2 −
p2 = 0√⇔ p = ±
2 ⇒ x8 + 1 = x4 + 2x2 + 1 x4 − 2x2 + 1 √


Ta câ : x4 + 2x2 + 1 = x2 + qx + 1 x2 −p qx + 1 = x4 + 2 − q 2 x2 + 1 √ √ 2 −p 2 çng nh§t 2 v¸ ta câ : 2 −p q 2 = 2 ⇔ q =±    √ 2 √ √ 4 2 2 ⇒ x + 2x + 1 = x + 2 − 2x + 1 x − 2 − 2x + 1 √


Ta câ : x4 − 2x2 + 1 = x2 + rx + 1 x2 − rx + 1 p= x4 + 2 − r2 x2 + 1 √ √ r =p ± 2+ 2  çng nh§t 2 v¸ ta câ ÷ñc : 2 − r2 = − 2⇔   p √ √ √ ⇒ x4 − 2x2 + 1 = x2 + 2 + 2x + 1 x2 − 2 + 2x + 1 √ Th½ dö 7 Do â m  ta câp:     p p p √ √ √ √ 2x + 1 x2 − 2 − 2x + 1 x2 + 2 + 2x + 1 x2 − 2 + 2x + 1 p p p √ √ √ 1 1 2 − 2x + 2 1 − 2 − 2x + 2 1 2 + 2x + 2 p p p = + + + ⇒ 8 x +1 8 x2 + 2 − √2x + 1 8 x2 − 2 − √2x + 1 8 x2 + 2 + √2x + 1 p √ 1 − 2 + 2x + 2 p + 8 x2 − 2 + √2x + 1 x8 +1 = x2 + 2− 8 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng Ta chó þ : Z 1 dx = ax2 + bx + c a Z dx , 4ac − b2 > 0 b c x2 + x + a a Z Z dx 1 dx 1 = (1) = p     2 2 2 2 a a (x − p) + q 2 −b 4ac − b + x− 2a 2a √ 4ac − b2 °t : p = −b ,q = , x = p + qt 2a 2a Z Z 1 1 2 1 2 Do â : (1) = aq t2 + 1 dt = √ dt = √ arctan t 2 t +1 4ac − b2 4ac − b2 2 2ax + b =√ arctan √ +C 4ac − b2 4ac − b2 Trong â ta °t : t = x −q p = √2ax + b 2 Z Z 4ac − b Z Z xdx 2ax + b − b 2ax + b 1 1 b L¤i câ : ax2 + bx + c = 2a ax2 + bx + c dx = 2a ax2 + bx + c dx− 2a ax2 +1bx + c dx Z  2  b 1 dx = ln ax + bx + c − +C 2a 2a ax2 + bx + c Vîi : aZ = c = 1, ∆ = 4 − b2 >Z 0 ta câ : Z xdx dx Ax + B dx = A +B 2 2 2 x + bx + 1 x + bx + 1 x + bx + 1  2x + b A  2 2B − Ab arctan √ + ln x + bx + 1 + C = √ 2 4 − b2 b2 p 4 −√ Z +2 Do â : A7.1 = 81 2 2p− 2x dx = √ x + 2 − 2x + 1 p p p √ √ √   p √ 2+ 2 2x + 2 − 2 2− 2 2 + = arctan p ln x + 2 − 2x + 1 + C1 √ 8 16 2 + 2 p √ Z 1 − 2 − 2x + 2 p dx = A7.2 = √ 8 2− x 2 − 2x + 1 p p p √ √ √   p √ 2+ 2 2x − 2 − 2 2− 2 2 = arctan p − ln x − 2 − 2x + 1 + C2 √ 8 16 2 + 2 p √ Z 2 + 2x + 2 p A7.3 = dx = √ 2+ x 2 + 2x + 1 p p p √ √ √   p √ 2− 2 2x + 2 − 2 2+ 2 2 = arctan p + ln x + 2 + 2x + 1 + C3 √ 8 16 2 − 2 p √ Z − 2 + 2x + 2 p A7.4 = dx = √ 2− x 2 + 2x + 1 p p p √ √ √   p √ 2− 2 2x − 2 − 2 2+ 2 2 p = arctan − ln x − 2 + 2x + 1 + C4 √ 8 16 2− 2 Do â m  ta câ : A7 = A7.1 + A7.2 + A7.3 + A7.4 C¡ch 2 : Ta câ : x8 + 1 = 0 ⇔ x8 = −1 = cos (π + k2π) + i sin (π + k2π) 9 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng     π + k2π π + k2π ⇒ x = cos + i sin , vîi : k = 0, .., 7 8 8 √ √ 3π 1p 5π π 1p 3π Ta câ : sin π8 = sin 7π = cos = = sin = cos = 2 − 2, sin 2+ 2 , 8 8 2 8 8 8 2 √ 7π 1p 5π = cos =− 2 − 2, cos 8 8 2      Z 3   dx 1X (2k + 1) π (2k + 1) π 2 Do â m  ta câ : 1 + x8 = − 8 ln x − 2x cos +1 ×cos + 8 8 k=0    (2k + 1) π   3 x sin (2k + 1) π 1 X   8 + + C, k = 0, ..., 7 arctan   × sin (2k + 1) π 4 8 k=0 1 − x cos 8 tr¶n ta nhªn th§y c¡c h» sè : α, β, χ, δ..., l  c¡c h» sè "b§t ành" tùc l  "nhúng h» Qua c¡c th½ dö sè khæng h· thay êi " dò ta câ thay êi "gi¡ trà cõa bi¸n x" . º t¼m hiºu rã v§n · n y ta s³ l m quen d¤ng to¡n n y vîi c¡cZb i tªp m¨u t÷ìng tü nh÷ sau : B i 1 : T¼m nguy¶n h m I1 = (x −x3)+(x27+ 3) dx Gi£ sû r¬ng : (x −x3)+(x27+ 3) = x −α 3 + x +β 3 ⇔ x + 27 = α (x + 3) (α + β) + (3α − 3β) ( + β (x − 3) = x( C¥n b¬ng h» sè 2 v¸ ta câ h» : α+β =1 ⇔ α=5 β = −4 α−β =9 Z Vªy : (x −x3)+(x27+ 3) dx = x −5 3 dx − x +4 3 dx = 5 ln |x − 3| − 4 ln |x + 3| + c Z : T¼m nguy¶n h m I2 = (x −8x3)−(x17+ 4) dx Gi£ sû r¬ng : (x −8x3)−(x17+ 4) = x −α 3 + x +β 4 = α (x(x+−4)3)+(xβ +(x4)− 3) = x (α(x+−β)3)+(x(4α+ −4) 3β) ( ( Z Z B i 2 º tø â ta câ h» : α+β =8 ⇔ α=1 4α − 3β = −17 β=7 Z Z Vªy : (x −8x3)−(x17+ 4) dx = x −1 3 dx + x +7 4 dx = ln |x − 3| + 7 ln |x + 4| + c Z − 26 : T¼m nguy¶n h m I3 = x2 7x dx − 6x − 16 − 26 7x − 26 α β (α + β) x + (2α − 8β) Gi£ sû r¬ng : x2 7x = = + = − 6x − 16 (x − ( 8) (x + 2) x − 8 x( +2 (x − 8) (x + 2) Z B i 3 C¥n b¬ng h» sè 2 v¸ m  ta câ h» : α+β =7 ⇔ α=3 2α − 8β = −26 β=4 Z − 26 3 4 Vªy : x2 7x dx = dx + dx = 3 ln |x − 8| + 4 ln |x + 2| + c − 6x − 16 x x+2 Z− 82 − 150 : T¼m nguy¶n h m I4 = 7x x+3 75x dx − 25x 2 − 150 7x2 + 75x − 150 α β χ Gi£ sû r¬ng : 7x x+3 75x = = + + − 25x x (x − 5) (x + 5) x x−5 x+5 Z Z B i 4 10 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng
C¥n b¬ng
α x2 − 25 + β x2 + 5x + χ x2 − 5x 7x2 + 75x − 150 = ⇔ x (x − 5) (x + 5) x (x − 5) (x + 5) (α + β + χ) x2 + (5β − 5χ) x − 25α = x (x − 5) (x +  5)     α+β+χ=7  α=6 h» sè 2 v¸ m  ta câ h» : 5β − 5χ = 75 ⇔ β = 8   −25α = −150 Z Z
  χ = −7 Z 7x2 + 75x − 150 6 8 7 dx = dx + dx − dx 3 x − 25x x x−5 x+5 =Z 6 ln |x| + 8 ln |x − 5| − 7 ln |x + 5| + c 2 − 49 : T¼m nguy¶n h m I5 = −2xx3−−14x dx 2 7x 2 − 49 −2x2 − 14x − 49 α β χ sû r¬ng : −2xx3−−14x = = + 2+ 2 2 7x x (x −
7) x x x−7 2 2 = α x − 7x + β (x −  7) + χx = (α + χ) x2 + (−7α + β) x − 7β  Vªy : Z B i 5 Gi£   α + χ = −2 C¥n b¬ng h» sè 2 v¸ ta câ h» :  −7α + β = −14 Vªy : Z ⇔  −7β = −49 Z Z −2x2 − 14x − 49 dx = x3 − 7x2 Z B i 6 : T¼m nguy¶n h m I6 = Gi£ sû r¬ng : 8x − 36 (x − 5) 2 = 3 dx + x 8x − 36 (x − 5)2 β 7 dx − x2   α=3 β=7   χ = −5 Z 5 7 dx = 3 ln |x| − − 5 ln |x − 7| + c x−7 x dx α α (x − 5) + β αx − 5α + β + = = 2 2 x − 5 (x − 5) (x − 5)( (x − 5)2 ( C¥n b¬ng h» sè 2 v¸ ta câ h» : α=8 α=8 ⇔ β=4 −5α + β = −36 Z 4 4 Vªy : 8x − 362 dx = x −8 5 dx + dx = 8 ln |x − 5| − +c 2 x−5 (x − 5) (x − 5) Z 2 + 20x + 9 : T¼m nguy¶n h m I7 = 6x dx x3 + 2x2 + x 2 + 20x + 9 6x2 + 20x + 9 α β χ Gi£ sû r¬ng : 6x = = + + x3 + 2x2 + x x + 1 (x + 1)2 x x(x + 1)2 2 αx (x + 1) + βx + χ(x + 1) (α + χ) x2 + (α + β + 2χ) x + χ = = 2 2 x(x + 1)   x(x + 1)    α+χ=6  α = −3 C¥n b¬ng h» sè 2 v¸ ta câ h» : α + β + 2χ = 20 ⇔ β = 5 Z Z B i 7 Vªy : Z 6x2 + 20x + 9 dx = − x3 + 2x2 + x   χ=9 Z 3 dx + x+1 Z 5   χ=9 Z 2 dx + (x + 1) 5 = −3 ln |x + 1| − + 9 ln |x| + 1 x+1 9 dx x 11 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng B i 8 : T¼m nguy¶n h m I8 = 9x3 − 20x2 + 30x − 97 dx (x − 2) (x − 3) (x2 + 5) 9x3 − 20x2 + 30x − 97 α β χx + δ Gi£ sû r¬ng : (x = + + − 2) (x − 3) (x2 + 5) x − 2 x − 3 x2 + 5 3 2 9x3 − 20x2 + 30x − 97 9x − 20x + 30x − 97 = 5 , β = lim =4 α = lim x→3 x→2 (x − 3) (x2 + 5) (x − 2) (x2 + 5) 9x3 − 20x2 + 30x − 97 +δ Do â : (x − 2) (x − 3) (x2 + 5) = x −5 2 + x −4 3 + χx 2 x +5 5 4 1 97 Cho x = 0 ta câ : − 30 = − 2 − 3 + 5 δ 1 13 x = 1 ta câ : − = −5 − 2 + (χ + δ) 2 (6 1 ( 3 δ= C¥n b¬ng c¡c h» sè 2 v¸ ta câ h» : 5 5 ⇔ χ = 0 δ=3 χ+δ =3 9x3 − 20x2 + 30x − 97 5 4 3 Do â ta câ : (x = + + − 2) (x − 3) (x2 + 5) Z x − 2 x −Z3 x2 + 5 Z Z 3 9x − 20x2 + 30x − 97 5 4 3 Vªy : (x dx = dx + dx + dx 2 2 − 2) (x − 3) (x + 5) x−2 x−3 x + 5  3 x = 5 ln |x − 2| + 4 ln |x − 3| + √ arctan √ +c 5 5 Z 3 2 : T¼m nguy¶n h m I9 = 4x − 21x 2 + 248x + 17 dx (x − 3) (x + 7) 3 2 +δ Ta gi£ sû r¬ng : 4x − 21x 2 + 248x + 17 = x −α 3 + β 2 + χx 2 x +7 (x − 3) (x + 7) (x − 3) Z B i 9 0 4x3 − 21x2 + 48x + 17 4x3 − 21x2 + 48x + 17 α = lim = 0 , β = lim =5 x→3 x→3 x2 + 7 x2 + 7 3 2 5 +δ χx + δ = Do â : 4x − 21x 2 + 248x + 17 = x −0 3 + 5 2 + χx + x2 + 7 x2 + 7 (x − 3) (x + 7) (x − 3) (x − 3)2 5 1 Cho x = 0 ta câ : 17 = + δ ⇔ δ = −2 63 9 7 3 5 1 x = 1 ta câ : = + (χ + δ) ⇔ χ + δ = 2 ⇔ χ = 2 − δ = 4 2 4 8 4x3 − 21x2 + 48x + 17 5 4x − 2 Suy ra : = + 2 2 2 2 + 7) x +7 (x − Z 3) Z (x3− 3) (x Z 2 4x − 21x + 48x + 17 4x − 2 5 Do â : dx = dx + dx 2 2 x2 + 7 (x − 3) (x2 + 7) (x − 3) Z Z Z 5 4x dx dx + dx − 2 = 2 2 2 x +7 x +7 (x − 3)  
5 2 x =− + 2 ln x2 + 7 − √ arctan √ +c x−3 7 7    Ta c¦n chó þ nhúng v§n · nh÷ sau : P (x) Gi£ sû ta câ h m ph¥n thùc : Q trong â bªc P (x) nhä hìn bªc cõa Q (x) v  ta gi£ sû r¬ng : (x) Q (x) = K (x) .R (x) = K (x) .(x − a)n 12 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng P (x) P (x) T (x) α β χ Do â : Q = + + ... + n = n + (x) K (x) (x − a) x−a K (x) .(x − a) (x − a)n−1 Trong â bªc cõa T (x) nhä hìn bªc  t¼m c¡ch h» sè α, β, ..., χ ta l m nh÷ sau :  cõa K (x).º 0 P (x) P (x) α = lim , β = lim x→a K (x) x→a K (x) P (x) , χ = lim x→a K (x) n Mët v§n · núa khi ph¥n t½ch mët ph¥n thùc l  c¡c ph¥n thùc m  khi m¨u câ nghi»m phùc .
P (x) Cho ph¥n thùc câ d¤ng : Q .Trong â : Q (x) = K (x) .R (x) = K (x) . ax2 + bx + c (x) 2 Vîi : ∆ = b − 4ac < 0 ta câ c¡c nghi»m nh÷ sau : xk = α ± βi, k = 0, 1, i2 = −1 P (x) P (x) T (x) χx + δ Do â ta câ : Q = = + 2 2 (x) K (x) . (ax + bx + c) K (x) ax + bx + c º t¼m c¡c h» sè : χ, δ ta l m nh÷ sau :   P (x) Im [ε + φi] Im [ε + φi] χ = lim , φ = Rec [ε + φi] + |α| . = β β x→α+βi K (x) C¡c th½ dö m¨u nh÷ sau : 2 +β χx + δ T¼m c¡c h» sè α, β, χ, δ : (x2 + 4xx+ +6) x(x+2 +1 2x + 6) = x2αx + 2 + 4x√+ 6 x + 2x + 6 2 Ta câ nghi»m cõa ph÷ìng tr¼nh : x + 4x + 6 = 0 ⇔ x = √ −2 ± 2 i 2 Nghi»m cõa  : x + 2x  ph÷ìng tr¼nh √ + 6 = 0 ⇔ x = −1 ± 5 i lim √ x→−2+ 2i 2 5 2 x2 + x + 1 = − i 2 x + 2x + 6 3 12 Do â ta câ : √  2 5 2 Im − i 3 12 5 √ • α= =− , 12 2  √  2 5 2  √  Im − i 3 12 2 10 2 2 5 2 √ − i + 2. = − =− • β = Rec 3 12 3 12 12 2  2  √ x +x+1 1 5 5 T÷ìng tü ta công câ : lim √ x2 + 4x + 6 = − 12 + 12 i x→−1+ 5i  √  1 5 5 i Im − + 12 12 5 √ • χ= = , 12 5  √  1 5 5  √  Im − + i 12 12 1 5 5 1 5 4 √ • δ = Rec − + i + =− + = 12 12 12 12 12 5  2 −2 5x + 4 Vªy ta câ ÷ñc : (x2 + 4xx+ +6) x(x+2 +1 2x + 6) = 12 (x−5x + 2 + 4x + 6) 12 (x2 + 2x + 6) 2 +β χ δ T¼m c¡c h» sè α, β, χ, δ : 2 x + 1 2 = αx + + 2 x +6 x + 2 (x + 2)2 (x + 6) (x + 2) √ Nghi»m cõa ph÷ìng tr¼nh : x2 + 6 = 0 ⇔ x = ± 6 i 13 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng x2 + 1  Ta câ : lim√ 2 x→ 6i (x + 2) Do â m  ta câ :  √ 6 1 + i = 10 5 √  √ 1 6 6 Im + i 10 5 1 √ • α= = √5 = 5 6 6  √  1 6  √  Im + i 10 5 1 6 1 √ • β = Rec = + i + 0. 10 5 10 6   • δ = lim x→−2 x2 + 1 4+1 5 1 = = = 2 x +6 4+6 10 2  • χ = lim x→−2  x2 + 1 x2 + 6 Vªy ta câ ÷ñc : 0  = lim x→−2 x2 + 1  10x (x2 + 6) 2 =− 20 1 =− 100 5 1 1 2x + 1 − + 2 (x2 + 6) (x + 2) Z 10 (x + 6) 5 (x + 2) 2(x + 2)2 : T¼m nguy¶n h m I10 = (x2 − x x++1)1(x2 + 1) dx +δ Gi£ sû r¬ng : (x2 − x x++1)1(x2 + 1) = x2αx− +x +β 1 + χx x2 + √1 Nghi»m cõa ph÷ìng tr¼nh : x2 − x + 1 = 0 ⇔ x = 21 ± 23 i h x + 1 i 3 √3 Ta câ : lim = − i √ 2 2 2 x→ 1 + 3 i x + 1 2 = B i 10 Do â m  ta câ : 2 2 √  3 3 − i Im 2 2 √ = −1 • α= 3 2  √  3 3 1 3 1 • β = Rec − i + |−1| = + = 2 2 2 2 2 2  Ta câ nghi»mh cõa ph÷ìngi tr¼nh : x2 + 1 = 0 ⇔ x = ±i x+1 L¤i câ : x→i lim 2 = −1 + i x −x+1 Do â suy ra : • χ = Im [−1 + i] = 1 , δ = Rec [−1 + i] = −1 +2 x−1 + 2 Do â ta câ : (x2 − x x++1)1(x2 + 1) = x2−x −x+1 x +1 Vªy : Z (x2 x+1 dx = − x + 1) (x2 + 1) 14 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng Z Z Z Z 2x − 1 3 dx 1 2x dx 1 dx + + dx − =− 2+1 2 x2 − x + 1
2 x2 − x + 1 2 x2 + 1
x Z Z Z Z 2 2 d x −x+1 d x +1 3 dx 1 dx 1 + + − =−   2 2 2 2 2 x −x+1 2 2 x +1 x +1 1 3 x− +  2 4 √

2x − 1 1 1 √ + ln x2 + 1 − arctan x + c = − ln x2 − x + 1 + 3 arctan 2  3  2   2 √ x +1 2x − 1 1 √ = ln + 3 arctan − arctan x + c 2 2 x −x+1 3 B i tªp b¤n åc tü luy»n : Z B i 1 : T¼m nguy¶n h m I1 = Z B i 2 : T¼m nguy¶n h m I2 = B i 3 √ x− 2 dx (x − 1) (x + 1) (x + 2) x2 − 1 dx
√ 2 − 3x + 2 (x2 − 1)2 x Z 1 dx : T¼m nguy¶n h m I3 = 2 2 2 Z (x + x2 + 2) (x + 1) x −1 : T¼m nguy¶n h m I4 = dx 3 (x + 2) (x − 1)3 B i 4 PH×ÌNG PHP L×ÑNG GIC HÂA TRONG TCH PH…N. D¤ng t½ch ph¥n êi bi¸n sè i·u ki»n bi¸ni sè h √
π π f x, a2 − x2 dx x = a sin t t∈ − ; h π  2  2 3π  √
a f x, x2 − a2 dx x= t ∈ 0; ∪ π; cos t 2h  2 √
π f x, x2 + a2 dx x = a tan t t ∈ 0;   r  2π  a+x dx f x, x = a cos 2t t ∈ 0; a−x  p  h π2 i f x, (x − a) (b − x) dx x = a + (b − a) sin2 t t ∈ 0; 2 B£ng 1: B£ng tâm t­t c¡c cæng thùc bi¸n êi l÷ìng gi¡c b¬ng c¡ch êi bi¸n sè. √ 1. D¤ng 1 : f x, a2 − x2 dx Z B i 1 : T¼m nguy¶n h m A1 = √1xdx − x2
°t : xZ = sin t ⇒ dx Z= cos tdt Z xdx sin t. cos tdt p Vªy : √ = = 2 2 1−x Z sin t. cos tdt |cos t| 1− Z sin t sin t. cos tdt π = sin tdt = − cos t + c, 0 ≤ t < ,  cos t 2 Z = Z sin t. cos tdt 3π − = − sin tdt = cos t + c, π ≤ t < , cos t 2 15 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng C¡ch 2 : i·u ki»n :Z−1 < x < 1, Z Bi¸n êi : √ xdx 2 = Z xdx √ √ 1 + x. 1 − x 1−x (1 − x) (1 + x) √ √ √ √ Ta nhªn th§y : 1 − x + 1 + x = 2. °t : 1 − x = 2 sin t, 1 + x = 2 cos t Do âZta câ : 1 − x = 2sin2t ⇒Zdx = −4 sin2t.
cos tdt Z 1 − 2sin t sin t. cos tdt xdx √ Vªy : √ = −2 cos 2tdt = − sin 2t + c = −4 2 sin t. cos t 1 + x. 1 − x C¡ch 3 :Z Ta câ : xdx = p
p
− 1
d 1 − x2 1 √ 1 − x2 2 d 1 − x2 = − 1 − x2 + c =− 2 Z1 − x2 dx √ : T¼m nguy¶n h m A2 = √ 2+x+ 2−x Ta chó þ : 2 + x√+ 2 − x = 4, √ Do â ta °t : 2 + x = 2 sin t, 2 − x = 2 cos t, 0 ≤ t ≤ π2 ; ⇒ 2 + x = 4sin2 t ⇒Z dx = 8 sin t cos tdtZ Z dx sin t cos tdt (sin t + cos t)2 − 1 √ Vªy : √ =4 =2 dt sin t + cos t sin t + cos t 2 + x + 2 − x Z Z Z Z √ dt dt   = 2 (sin t + cos t)dt − 2 = 2 (sin t + cos t)dt − 2 π sin t + cos t sin t + 4 Z Z √ dt = 2 (sin t + cos t)dt − 2 t π t π 2 sin + . cos + 2 8 2 8   √ t π   = 2 (sin t − cos t) − 2 ln tan + +c 2 8 √ B i 2 xdx 1 √ =− 2 1 − x2 Z Z 3 B i 3 : T½nh t½ch ph¥n A3 = Z2 1 − x2
p 1 − x2 dx 0 °t : x = sin t ⇒ dx = cos tdt. êi cªn : Vîi x = 0 ⇒ t = 0, x = π Vªy : A3 = π Z3
p 1 − sin2 t 1 − sin2 t cos tdt = 0 Z 1 (1 + cos 2t) dt = 4 2 Z 1 + 2 cos 2t + cos2 2t dt
0 Z3  π 1 + 2 cos 2t +  1 + cos 4t 2 0    1  = 3t + 2 sin 2t + sin 4t  8 4   1 C¡ch 2 : cos4 tdt 0 π 1 4 Z3 π 3 0 = 3 π ⇒t= 2 3 π 1 − sin2 t cos2 tdt =
0 π 3 1 = 4 Z3 √ dt = 1 8 Z3 (3 + 4 cos 2t + cos 4t) dt 0 π 3 0 √ π 7 3 = + 8 64 16 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng √ √ √ Vîi i·u ki»n : 0 ≤ x ≤ 23 ta câ : 1 − x2 = 1 − x. 1 + x Chó þ r¬ng : 1 √ −x+1+x=2 √ √ √ Do â ta °t : 1 − x = 2 sin√t, 1 + x = 2 cos t, dx = −4 sin t. cos tdt π êi cªn : x = 0 ⇒ t = π4 , x = 23 ⇒ t = 12 p Vªy : A3 = −32 π π Z12 Z12 sin2 t.cos2 t. sin t. cos t. sin t. cos tdt = −32 π 4 π 12 = −2 4 sin 2tdt = −2 π 4 Z   1 − cos 4t 2 2 π 4 B i 4 : T½nh t½ch ph¥n A4 = π Vªy : A4 =   1  = tan t  2  π 4 = 0 π 12 π 4 1 + cos 8t dt 2  √ π 7 3 = + 8 64 π Z4
p = 2 2 − 2sin t 2 − 2sin t 0 1 − 2 cos 4t + êi cªn : x = 0 ⇒ t = 0, x = 1 ⇒ t = π4 , √ 2 cos tdt 2 Z  dx √ (2 − x2 ) 2 − x2 √ √ 0 x = 2 sin t ⇒ dx = 2 cos tdt. Z4 1 dt = − 2 π 4     1 3 1 1  =− t − sin 4t + sin 8t  2 2 2 16  Z1 °t : π 4 π 12 π 12 Z sin4 t.cos4 tdt π √ 1 2 cos tdt
√ = 2 2 2 − 2sin t 2 cos t 0 Z4 dt cos2 t 0 1 2 C¡ch 2 : q √ p√ p
√
p√ 2 Vîi i·u ki»n : 0 ≤ x ≤ 1 ta câ : 2 − x = 2−x 2+x = 2 − x. 2+x q √ q p p √ √ √

°t : 2 − x = 2 2 sin t, 2 + x = 2 2 cos t, √ √ √ Do â : 2 − x = 2 2 sin t, dx = −4 2 sin t. cos tdt êi cªn : x = 0 ⇒ t = π4 , x = 1 ⇒ t = π8 , π Vªy : A4 = Z8 π 4 π π √ Z8 Z8   −4 2 sin t. cos tdt dt 1 1 1 √ = − = − 1 + d (tan t) 4 4 tan2 t sin2 t.cos2 t 8sin2 t.cos2 t.2 2 sin t. cos t π 4    1 1  =− tan t −  4 tan t  Z0 B i 5 : T½nh t½ch ph¥n A5 = π 8 = π 4 π 4 1 1 −0= 2 2 dx 1+ p −x (x + 1) − 12 Ta câ : −x (x + 1) = −
1 1 1 x2 + x = − x2 + 2. .x + − 2 4 4   = 1 1 − x+ 4 2  2 17 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng °t : x + 12 = 21 sin t ⇒ dx = 12 cos tdt. êi cªn : x = − 21 ⇒ t = 0, x = 0 ⇒ t = π2 , Z0 Vªy : A5 = − 12 = π 2 = π 2 π π 1 2 2 Z Z cos tdt cos tdt dt dx π 2 r r = = −2 =   2 + cos t 2 2 + cos t 1 1 2 1 1 2 − sin t 0 1+ 0 0 1+ − x+ 4 2  4 t 4  π π t  π 2 2 Z Z d tan  2 tan dt π π 4  2 2   √ −2 = −4 = − √ arctan  t t 2 2 0 3 3 2 2 1 + 2cos tan + 3 0 0  2 2  √ 0 1 4 2π 9−4 3 4 π + √ arctan √ π − √ arctan √ = − √ = 2 3 3 18 3 3 3 3 π Z2 C¡ch 2 : √ √ Ta chó þ : −x + x + 1 = 1 . °t : −x = sin t, 1 + x = cos t 1 π ⇒ x = −sin2 t ⇒ dx = −2 sin t. cos tdt. êi cªn : x = − ⇒ t = , x = 0 ⇒ t = 0 2 4 Vªy : Z0 Z0 dx 1+ p −x (x + 1) − 21 = π −2 2 = −2 sin t. cos tdt =− 1 + sin t. cos t Z0 (2 + 2 sin t. cos t − 2) dt 1 + sin t. cos t π 4 π 4 π π Z4 Z4 π dt = −2 1 + sin t. cos t 2 0 d (tan t) tan2 t + tan t + 1 0 π = π −2 2 Z4  0 °t : u = tan t, Vîi : π π d (tan t) = − 2J  2 2 1 3 tan t + + 2 4 π t = 0 ⇒ u = 0, t = ⇒ u = 1 4 Z1 Z4 d (tan t) dt ⇒J = =     2 1 1 2 3 3 tan t + t+ + + 0 0 4 2 4 √ 2 √
1 3 3 °t : t + 2 = 2 tan v ⇒ dt = 2 1 + tan2v dv êi cªn : t = 0 ⇒ v = π6 , t = 1 ⇒ v = π3 , √ π π  Z1 Z3 3 1 + tan2 v
dv Z3  dt 2 2  2 Do â :  = =√ dv = √ v   3 3 1 2 3 2 3 3  tan v + π π t+ + 0 4 4 6 6 2 4 √ π π 2π 9−4 3 Vªy ta câ ÷ñc : I = 2 − 2J = 2 − √ = 18 π 3 3 √ π 3 π 6 π = √ 3 3 2. D¤ng 2 : f x, x2 − a2 dx Z p B i 1 : T¼m nguy¶n h m B1 = x x2 − 1 dx
18 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng °t : x = cos1 t , t ∈ Vªy : h π h ∪ π; B i 2 : T½nh t½ch ph¥n B2 = °t : Vªy : Z4 3π 2 , sin t dt 2 cos2 t   r r  sin t  2 1 sin t  Z Z Z sin t.  −1 cos t  sin t cos2 t cos2 t sin t B1 = . dt = . dt = dt cos t Z cos2 t cos t cos2 t cos3 t Z Z sin2 t 1 − cos2 t 1 − cos2 t = dt = dt = d (tan t) 4t 4t 2t cos cos cos Z
tan3 t = 1 + tan2 t − 1 d (tan t) = +c 3 0; x2  ⇒ dx = p x2 − 4 dx 2 2 sin t π 2 ⇒ dx = dt , êi cªn : Vîi gi¡ trà x = 2 ⇒ t = 0, x = 4 ⇒ t = x= cos t r cos2 t 3 r 2 π π π 4 Z3 4 Z3 4 sin t Z3 −4 2 2 sin t sin2 t cos t cos2 t 2 sin t B2 = . dt = 16 . dt = 16 dt cos2 t cos2 t cos2 t cos2 t cos5 t 0 0 0 π π Z3 Z3 sin2 t 0 1 − sin2 t √ = 16 sin2 t d (sin t) = 16 cos6 t 0 L¤i °t : u = sin t, t = 0 ⇒ u = 0, t = π3 ⇒ u = Do â : sin2 t (1 − u2 ) 16u2 (2) 3 u2 =− 3 2 (u − 1)3 (1 + u)3 α β χ δ ε φ 16u2 = + + + + + − 3 3 2 3 2 u − 1 (u − 1) 1 + u (1 + u) (1 − u) (1 + u) (u − 1) (1 + u)3 1 − sin2 t Ph¥n t½ch : u2
3 =
3 d (sin t)  • α = lim − x→1 • δ = lim = 1, β = lim − (1 + u)3 16u2   x→1 (2) 16u2 , (1 + u)3  16u2  = −1, χ = lim − = −1, φ = lim − 3 x→−1 (u − 1) (u − 1)3   16u2 χ = lim − =2 x→−1 (u − 1)3 x→−1 Suy ra : − 16u2 3 (1 − u) (1 + u) π 3 Vªy : 16 − sin2 t Z 0 3 = x→1 16u2 (1 + u)3  = −2 , = −1, 1 1 2 1 1 2 − − − − + 2 3 2 u − 1 (u − 1) (1 + u) (1 + u) (u − 1) (1 + u)3
3 d (sin t) = 1 − sin2 t √ 3 2 Z  =  1 1 2 1 1 2 − − − − + du 2 3 2 u − 1 (u − 1) (1 + u) (1 + u) (u − 1) (1 + u)3 0 19 Tuyºn tªp c¡c chuy¶n · v  k¾ thuªt t½nh nguy¶n h m,t½ch ph¥n. Nguy¹n Duy «ng √ 3 Z2 " u2 = −16 (u − 1)3 (1 + u)3 du = 0 #    u − 1    + ln    2 2 u + 1  (u − 1) 2 u3 + u
√ 3 2 √ = 28 3+ln 0  √  2− 3 √ 2+ 3 C¡ch 2 : p + t) °t : t = x2 − 4 ⇒ t2 = x2 − 4 ⇒ tdt = xdx ⇒ dxt = d (x x+t Do c¡ch °t m  ta câ : t2 = x2 − 4 ⇔ (t − x) (t + x) = −4 Ta l¤i °t : u = t + x ⇒ t − x = − u4 √ êi cªnZ : x = 2 ⇒ t = 2, x =Z4 ⇒ t=4+2 3  2  2 du Z  16 u3 2  p 1 4 4 2 − Vªy : x x2 − 4dx = 16 u + u u − u u = + du u5 16 4 4 u4 − 2 ln |u| + c + 4 u 64    4 u4  B2 = − 4 + − 2 ln |u|  u 64  =− Thay cªn v o ta câ : B i 3 : T½nh t½ch ph¥n B3 = Z2 √ 4+2 3 √ √
= 28 3 − 2 ln 2 + 3 2 dx √ (x2 − 1) x2 − 1 √2 3 tdt 2 .êi cªn : Vîi gi¡ trà x= √ °t : x = cos1 t ⇒ dx = sin 2 cos t π 3 Vªy : B3 = Z π 6 3 π 3 Z sin tdt cos2 t   1  =−  sin t   1 cos2 t π 3 π 6 r 1 −1 √ 6−2 3 = 3 cos2 t = −1 cos tdt = sin2 t π 6 ⇒t= π π ,x = 2 ⇒ t = 6 3 π Z3 d (sin t) sin2 t π 6 C¡ch 2 : p °t : t = x2 − 1 ⇒ t2 = x2 − 1 ⇒ tdt = xdx, dtx = dxt dt d (x + t) Do â ta suy ra : dxt = dxx + = +t x+t Do c¡ch °t m  ta câ : t2 = x2 − 1 ⇒ (t + x) (t − x) = −1 √ √ °t : u = t + x ⇒ t − x = − u1 ,Vîi gi¡ trà x = √2 ⇒ u = 3, x = 2 ⇒ u = 2 + 3 Vªy : √ 2+ Z 3 √ 2+ Z 3 3 du udu B3 = 4 =4 =2   2 2 − 1)2 1 (u √ √ u− u 3 3  √u √  2+ 3 2  6−2 3 = =  1 − u2  √3 3 B i 4 : T½nh t½ch ph¥n B4 = Z0 √ 2+ Z 3
d u2 − 1 2 √ 3 (u2 − 1) dx √ (x + 1) 3 − 2x − x2 − 12 20