How do you evaluate the integral $int lnx/(1+x)^2$ ?

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Answer 1 · 2017-01-19T20:47:08

$-lnx/(1+x)+lnx-ln(1+x)+C$

$I=intlnx/(1+x)^2dx$

Integration by parts takes the form $intudv=uv-intvdu$ . Let:

$u=lnx$
$dv=1/(1+x)^2dx$

Differentiate $u$ and integrate $dv$ . The integration of $dv$ is best performed with the substitution $t=1+x=>dt=dx$ . You should get:

$du=1/xdx$
$v=-1/(1+x)$

Then:

$I=uv-intvdu$

$I=-lnx/(1+x)-int1/x(-1/(1+x))dx$

$I=-lnx/(1+x)+int1/(x(1+x))dx$

Perform partial fraction decomposition on $1/(x(1+x))$ :

$1/(x(1+x))=A/x+B/(1+x)$

Then:

$1=A(1+x)+Bx$

Letting $x=-1$ :

$1=A(1-1)+B(-1)$

$B=-1$

Letting $x=0$ :

$1=A(1+0)+B(0)$

$1=A$

Then:

$1/(x(1+x))=1/x-1/(1+x)$

So:

$I=-lnx/(1+x)+int1/xdx-int1/(1+x)dx$

These are simple integrals. The second can be performed with the substitution $s=1+x=>ds=dx$ .

$I=-lnx/(1+x)+lnx-ln(1+x)+C$

How do you evaluate the integral int lnx/(1+x)^2int lnx/(1+x)^2?