# How do you evaluate the integral int lnx/(1+x)^2?

Jan 19, 2017

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

#### Explanation:

$I = \int \ln \frac{x}{1 + x} ^ 2 \mathrm{dx}$

Integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. Let:

$u = \ln x$
$\mathrm{dv} = \frac{1}{1 + x} ^ 2 \mathrm{dx}$

Differentiate $u$ and integrate $\mathrm{dv}$. The integration of $\mathrm{dv}$ is best performed with the substitution $t = 1 + x \implies \mathrm{dt} = \mathrm{dx}$. You should get:

$\mathrm{du} = \frac{1}{x} \mathrm{dx}$
$v = - \frac{1}{1 + x}$

Then:

$I = u v - \int v \mathrm{du}$

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

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

Perform partial fraction decomposition on $\frac{1}{x \left(1 + x\right)}$:

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

Then:

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

Letting $x = - 1$:

$1 = A \left(1 - 1\right) + B \left(- 1\right)$

$B = - 1$

Letting $x = 0$:

$1 = A \left(1 + 0\right) + B \left(0\right)$

$1 = A$

Then:

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

So:

$I = - \ln \frac{x}{1 + x} + \int \frac{1}{x} \mathrm{dx} - \int \frac{1}{1 + x} \mathrm{dx}$

These are simple integrals. The second can be performed with the substitution $s = 1 + x \implies \mathrm{ds} = \mathrm{dx}$.

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