what is the integral of (ln(x+1)/(x^2)) dx?

Answer 1

Use integration by parts
Let u = ln(x+1) and dv = 1/x^2 dx, then
du = 1/(x+1) dx and v = -1/x
uv - v du
= -(1/x) ln(x+1) + integral of [1/(x(x+1)) dx]
Use partial fraction decomposition to integrate the above integral. I will leave that to you.
Answer: -(1/x) ln(x+1) + ln(x) - ln(x+1) + C