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Question Number 86983 by mathmax by abdo last updated on 01/Apr/20
1) calculate ∫ (dx/((x+1)^3 (x−2)^3 )) 2) decompose the fraction F(x)=(1/((x+1)^3 (x−2)^3 ))
1)calculate∫dx(x+1)3(x−2)32)decomposethefractionF(x)=1(x+1)3(x−2)3
Commented by hknkrc46 last updated on 01/Apr/20
(1/((x+1)^3 (x−2)^3 )) =(1/(27))∙(([(x+1)−(x−2)]^3 )/((x+1)^3 (x−2)^3 )) =(1/(27))[(((x+1)^3 −(x−2)^3 −3(x+1)(x−2)[(x+1)−(x−2)])/((x+1)^3 (x−2)^3 ))] =(1/(27))[(1/((x−2)^3 ))−(1/((x+1)^3 ))−(9/((x+1)^2 (x−2)^2 ))] =(1/(27))[(1/((x−2)^3 ))−(1/((x+1)^3 ))−(([(x+1)−(x−2)]^2 )/((x+1)^2 (x−2)^2 ))] =(1/(27))[(1/((x−2)^3 ))−(1/((x+1)^3 ))−(((x+1)^2 +(x−2)^2 −2(x+1)(x−2))/((x+1)^2 (x−2)^2 ))] =(1/(27))[(1/((x−2)^3 ))−(1/((x+1)^3 ))−(1/((x−2)^2 ))−(1/((x+1)^2 ))+(2/((x+1)(x−2)))] =(1/(27))[(1/((x−2)^3 ))−(1/((x+1)^3 ))−(1/((x−2)^2 ))−(1/((x+1)^2 ))+(2/3)∙(((x+1)−(x−2))/((x+1)(x−2)))] =(1/(27))[(1/((x−2)^3 ))−(1/((x+1)^3 ))−(1/((x−2)^2 ))−(1/((x+1)^2 ))+(2/3)∙((1/(x−2))−(1/(x+1)))] =(1/(27))[(1/((x−2)^3 ))−(1/((x+1)^3 ))−(1/((x−2)^2 ))−(1/((x+1)^2 ))]+(2/(81))∙((1/(x−2))−(1/(x+1))) ∫ (dx/((x+1)^3 (x−2)^3 ))=(1/(27))[(((x−2)^(−2) )/(−2))+(((x+1)^(−2) )/2)+(x−2)^(−1) +(x+1)^(−1) ]+(2/(81))ln(((x−2)/(x+1)))+c
1(x+1)3(x−2)3=127⋅[(x+1)−(x−2)]3(x+1)3(x−2)3=127[(x+1)3−(x−2)3−3(x+1)(x−2)[(x+1)−(x−2)](x+1)3(x−2)3]=127[1(x−2)3−1(x+1)3−9(x+1)2(x−2)2]=127[1(x−2)3−1(x+1)3−[(x+1)−(x−2)]2(x+1)2(x−2)2]=127[1(x−2)3−1(x+1)3−(x+1)2+(x−2)2−2(x+1)(x−2)(x+1)2(x−2)2]=127[1(x−2)3−1(x+1)3−1(x−2)2−1(x+1)2+2(x+1)(x−2)]=127[1(x−2)3−1(x+1)3−1(x−2)2−1(x+1)2+23⋅(x+1)−(x−2)(x+1)(x−2)]=127[1(x−2)3−1(x+1)3−1(x−2)2−1(x+1)2+23⋅(1x−2−1x+1)]=127[1(x−2)3−1(x+1)3−1(x−2)2−1(x+1)2]+281⋅(1x−2−1x+1)∫dx(x+1)3(x−2)3=127[(x−2)−2−2+(x+1)−22+(x−2)−1+(x+1)−1]+281ln(x−2x+1)+c
Commented by mathmax by abdo last updated on 01/Apr/20
thank you sir its a eazy way...
thankyousiritsaeazyway…
Commented by mathmax by abdo last updated on 01/Apr/20
1) I =∫ (dx/((x+1)^3 (x−2)^3 )) ⇒ I =∫ (dx/((((x+1)/(x−2)))^3 (x−2)^6 )) changement ((x+1)/(x−2)) =t give x+1 =tx−2t ⇒(1−t)x =−1−2t ⇒x =((2t+1)/(t−1)) ⇒ (dx/dt) =((2(t−1)−(2t+1))/((t−1)^2 )) =((−3)/((t−1)^2 )) also x−2 =((2t+1)/(t−1))−2=((2t+1−2t+2)/(t−1)) =(3/(t−1)) ⇒ I = ∫ ((−3dt)/((t−1)^2 t^3 ((3/(t−1)))^6 )) =−(1/3^5 ) ∫ (((t−1)^4 dt)/t^3 ) =−(1/3^5 )∫ ((Σ_(k=0) ^4 C_4 ^k t^k (−1)^(4−k) )/t^3 )dt =−(1/3^5 ) Σ_(k=0) ^4 (−1)^k C_4 ^k ∫ t^(k−3) dt =−(1/3^5 )Σ_(k=0 and k≠2) ^4 (−1)^k C_4 ^k (1/(k−2))t^(k−2) −(1/3^5 ) C_4 ^2 ln∣t∣ +C =−(1/3^5 ) Σ_(k=0 and k≠2) ^4 (((−1)^k C_4 ^k )/(k−2)) (((x+1)/(x−2)))^(k−2) −(1/3^5 ) C_4 ^2 ln∣((x+1)/(x−2))∣ +C −3^5 I =(C_4 ^0 /(−2))(((x+1)/(x−2)))^(−2) +(C_4 ^1 /1)(((x+1)/(x−2)))^(−1) −C_4 ^3 (((x+1)/(x−2))) +(C_4 ^4 /2)(((x+1)/(x−2)))^2 + C_4 ^2 ln∣((x+1)/(x−2))∣ +C
1)I=∫dx(x+1)3(x−2)3⇒I=∫dx(x+1x−2)3(x−2)6changementx+1x−2=tgivex+1=tx−2t⇒(1−t)x=−1−2t⇒x=2t+1t−1⇒dxdt=2(t−1)−(2t+1)(t−1)2=−3(t−1)2alsox−2=2t+1t−1−2=2t+1−2t+2t−1=3t−1⇒I=∫−3dt(t−1)2t3(3t−1)6=−135∫(t−1)4dtt3=−135∫∑k=04C4ktk(−1)4−kt3dt=−135∑k=04(−1)kC4k∫tk−3dt=−135∑k=0andk≠24(−1)kC4k1k−2tk−2−135C42ln∣t∣+C=−135∑k=0andk≠24(−1)kC4kk−2(x+1x−2)k−2−135C42ln∣x+1x−2∣+C−35I=C40−2(x+1x−2)−2+C411(x+1x−2)−1−C43(x+1x−2)+C442(x+1x−2)2+C42ln∣x+1x−2∣+C
Commented by mathmax by abdo last updated on 01/Apr/20
2) ∫ F(x)dx is known we use F(x) =(d/dx)(∫ F(x)dx)
2)∫F(x)dxisknownweuseF(x)=ddx(∫F(x)dx)
Answered by mind is power last updated on 02/Apr/20
F(a,b)∫(dx/((x+a)(x+b)))=(1/(b−a))ln(((x+a)/(x+b)))+c ∫(dx/((x+1)^3 (x−2)^3 ))=(∂^3 F/(4∂^2 a.∂^2 b))∣_(a=1,b.−2)
F(a,b)∫dx(x+a)(x+b)=1b−aln(x+ax+b)+c∫dx(x+1)3(x−2)3=∂3F4∂2a.∂2b∣a=1,b.−2

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