How to Integrate ln(x+1)/x^2 using Partial Fractions

[nameVoid]

$$\int \frac{ln(x+1)}{x^2}dx$$
$$u=x+1$$
$$\int\frac{lnu}{(u-1)^2}du$$
$$\int \frac{A}{u-1}+\frac{B}{(u-1)^2}du$$
$$lnu=A(u-1)+B$$
$$B=0,A=ln2$$
$$\int \frac{ln2}{u-1}du$$
$$ln2ln|x|+C$$

 

[rl.bhat]

You have to do integration by parts.
Int[ln(x+1)/x^2] = ln(x+1)*intg(1/x^2) - Intg{[Intg(1/x^2)*d/dx[ln(x+1)]}
Now proceed.

 

[nameVoid]

where is the error

 

[rl.bhat]

I = -1/x*ln(x+1) - int[(-1/x)*1/(x+1)]
= -1/x*ln(x+1) + int(1/x) - int1/(x+1)
Now find the integration.
In your partial factor, there is no ln function in the right hand side. So you can't equate the coefficients.