# How do you find the integral (ln(x+1)/(x^2)) dx?

Aug 6, 2015

$\left(- \frac{1}{x}\right) \ln \left(x + 1\right) + \ln x - \ln \left(x + 1\right)$ +C

#### Explanation:

Integrate the function by parts, being a product of functions $\frac{1}{x} ^ 2$ and ln (x+1)

The integral would be $\ln \left(x + 1\right) \left(- \frac{1}{x}\right)$ +$\int \frac{1}{x \left(x + 1\right)} \mathrm{dx}$

$\left(- \frac{1}{x}\right) \ln \left(x + 1\right) + \int \left(\frac{1}{x} - \frac{1}{x + 1}\right) \mathrm{dx}$

$\left(- \frac{1}{x}\right) \ln \left(x + 1\right) + \ln x - \ln \left(x + 1\right)$ +C