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Question Number 174902 by daus last updated on 14/Aug/22

Answered by Ar Brandon last updated on 14/Aug/22

∫_0 ^4 (1/((1+(√x))^2 ))dx , x=t^2 ⇒dx=2tdt =∫_0 ^2 ((2t)/((1+t)^2 ))dt=2∫_0 ^2 ((1/(1+t))−(1/((1+t)^2 )))dt =2[ln(1+t)+(1/(1+t))]_0 ^2 =2(ln3+(1/3)−1) =2(ln3−(2/3))=ln9−(4/3)

∫041(1+x)2dx,x=t2⇒dx=2tdt=∫022t(1+t)2dt=2∫02(11+t−1(1+t)2)dt=2[ln(1+t)+11+t]02=2(ln3+13−1)=2(ln3−23)=ln9−43

Commented by floor(10²Eta[1]) last updated on 14/Aug/22

it′s 2(ln3+(1/3))−2

it′s2(ln3+13)−2

Commented by Ar Brandon last updated on 14/Aug/22

Yes, you're right. Thanks!

Commented by daus last updated on 14/Aug/22

how to do partial fraction at (1/((1+t)^2 ))?

howtodopartialfractionat1(1+t)2?

Commented by Ar Brandon last updated on 14/Aug/22

(t/((1+t)^2 ))=(((1+t)−1)/((1+t)^2 ))=((1+t)/((1+t)^2 ))−(1/((1+t)^2 ))=(1/(1+t))−(1/((1+t)^2 ))

t(1+t)2=(1+t)−1(1+t)2=1+t(1+t)2−1(1+t)2=11+t−1(1+t)2

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