Question

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

Answer 1

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

Explanation:

`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`