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Question Number 85641 by abdomathmax last updated on 23/Mar/20
calculate A_λ =∫_3 ^∞ (dx/((x+λ)^3 (x−2)^4 )) (λ>0)
calculateAλ=∫3∞dx(x+λ)3(x−2)4(λ>0)
Commented by mathmax by abdo last updated on 24/Mar/20
A_λ =∫_3 ^(+∞) (dx/((((x+λ)/(x−2)))^3 (x−2)^7 )) changement ((x+λ)/(x−2))=t give x+λ =tx−2t ⇒(1−t)x=−2t−λ ⇒x=((−2t−λ)/(1−t)) =((2t+λ)/(t−1)) ⇒ (dx/dt) =((2(t−1)−(2t+λ))/((t−1)^2 )) =((2t−2−2t−λ)/((t−1)^2 )) =((−2−λ)/((t−1)^2 )) x−2 =((2t+λ)/(t−1))−2 =((2t+λ−2t+2)/(t−1)) =((λ+2)/(t−1)) ⇒ A_λ =− ∫_(3+λ) ^1 (1/(t^3 (((λ+2)/(t−1)))^7 ))×((λ+2)/((t−1)^2 ))dt =(1/((λ+2)^6 ))∫_1 ^(3+λ) (dt/(t^3 (((λ+2)/(t−1)))^(−5) )) =(1/((λ+2)))∫_1 ^(3+λ) (((t−1)^5 )/t^3 )dt ⇒ (λ+2)A_λ =∫_1 ^(3+λ) ((Σ_(k=0) ^5 C_5 ^k t^k (−1)^(5−k) )/t^3 )dt =−∫_1 ^(3+λ) Σ_(k=0) ^5 (−1)^k C_5 ^k t^(k−3) dt =−Σ_(k=0) ^5 (−1)^k C_5 ^k ∫_1 ^(3+λ) t^(k−3) dt =−Σ_(k=0 and k≠2) ^5 (−1)^k C_5 ^k [(1/(k−2))t^(k−2) ]_1 ^(3+λ) −C_5 ^2 ∫_1 ^(3+λ) (dt/t) =−Σ_(k=0) ^5 (((−1)^k C_5 ^k )/(k−2)){ (3+λ)^(k−2) −1}−C_5 ^2 {ln∣3+λ∣} ⇒ A_λ =−(1/(λ+2))Σ_(k=0and k≠2) ^5 (((−1)^k C_5 ^k )/(k−2)){(3+λ)^(k−2) −1} −(1/(λ+2)) C_5 ^2 ln∣3+λ∣ .
Aλ=∫3+∞dx(x+λx−2)3(x−2)7changementx+λx−2=tgivex+λ=tx−2t⇒(1−t)x=−2t−λ⇒x=−2t−λ1−t=2t+λt−1⇒dxdt=2(t−1)−(2t+λ)(t−1)2=2t−2−2t−λ(t−1)2=−2−λ(t−1)2x−2=2t+λt−1−2=2t+λ−2t+2t−1=λ+2t−1⇒Aλ=−∫3+λ11t3(λ+2t−1)7×λ+2(t−1)2dt=1(λ+2)6∫13+λdtt3(λ+2t−1)−5=1(λ+2)∫13+λ(t−1)5t3dt⇒(λ+2)Aλ=∫13+λ∑k=05C5ktk(−1)5−kt3dt=−∫13+λ∑k=05(−1)kC5ktk−3dt=−∑k=05(−1)kC5k∫13+λtk−3dt=−∑k=0andk≠25(−1)kC5k[1k−2tk−2]13+λ−C52∫13+λdtt=−∑k=05(−1)kC5kk−2{(3+λ)k−2−1}−C52{ln∣3+λ∣}⇒Aλ=−1λ+2∑k=0andk≠25(−1)kC5kk−2{(3+λ)k−2−1}−1λ+2C52ln∣3+λ∣.
Commented by mathmax by abdo last updated on 24/Mar/20
λ≠−2
λ≠−2

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