Question

What is `int (ln(x+1)/(x^2)) dx`?

Answer 1

`intln(x+1)/x^2dx=ln(x/(x+1))-ln(x+1)/x+"c"`

Explanation:

For `intln(x+1)/x^2dx`, we integrate by parts

Let `u=ln(x+1)rArrdu=1/(x+1)dx`

and `dv=1/x^2dxrArrv=-1/x`

So

`intln(x+1)/x^2dx=-ln(x+1)/x+int1/(x(x+1))dx=int1/x-1/(x+1)dx-ln(x+1)/x=lnx-ln(x+1)-ln(x+1)/x`