Question

How do you integrate `ln(x+1)/x^2`?

Answer 1

`-ln(x+1)/x+ln|x/(x+1)|+C, or,`

`ln{|x*(x+1)^(-(1/x+1))|}+C`

Explanation:

The Rule of Integration by Parts :

`intuv'dx=uv-intu'vdx`.

Let, `I=intln(x+1)/x^2dx`.

We take, `u=ln(x+1) and v'=1/x^2`.

`:. u'=1/(x+1)and v=intv'dx=int1/x^2dx=-1/x`.

`:. I=-1/xln(x+1)-int{(1/(x+1))(-1/x)}dx`,

`=-ln(x+1)/x+int1/{x(x+1)}dx`,

`=-ln(x+1)/x+int{(x+1)-x}/{x(x+1)}dx`,

`=-ln(x+1)/x+int{(x+1)/{x(x+1)}-x/{x(x+1)}}dx`,

`=-ln(x+1)/x+int{1/x-1/(x+1)}dx`,

`=-ln(x+1)/x+{ln|x|-ln|(x+1)|}`,

`=-ln(x+1)/x+ln|x/(x+1)|`.

`rArr I=-ln(x+1)/x+ln|x/(x+1)|+C,`

`or, I=-1/xln|x+1|+ln|x/(x+1)|`,

`=ln(x+1)^(-1/x)+ln|x/(x+1)|`,

`=ln{|(x+1)^(-1/x)*x/(x+1)|}`.

`:. I=ln{|x*(x+1)^(-(1/x+1))|}+C`.

Enjoy Maths.!

Answer 2

`-log(x+1)/x+log|x|-log|x+1|+C`

Explanation:

We use Integration by parts:
`int fdg=fg-intgdf`

with
`f=log(x+1),dg=1/x^2dx`
we get
`df=1/(x+1)dx,g=-1/x`
so we have
`-log(x+1)/x+int1/(x*(x+1))dx`
`=log(x+1)/x+int1/xdx-int1/(x+1)dx`
so we get
`-log(x+1)/x+log|x|-log|x+1|+C`