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

Jul 23, 2016

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

#### Explanation:

$\int \setminus \left(\ln \frac{x + 1}{{x}^{2}}\right) \setminus \mathrm{dx}$

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

by IBP this becomes:

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

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

so some partial fractions on this integral

$\frac{1}{x \left(x + 1\right)} = \frac{A}{x} + \frac{B}{x + 1} = \frac{A \left(x + 1\right) + B x}{x \left(x + 1\right)}$

$x = 0 , 1 = A$
$x = - 1 , 1 = - B$

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

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

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