# What is the antiderivative of  (ln(x+1)/(x^2))?

Jan 9, 2016

$I = - \ln \frac{x + 1}{x} + \ln | x | - \ln | x + 1 | + c$

#### Explanation:

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

Let's say $u = \ln \left(x + 1\right)$ so $\mathrm{du} = \frac{1}{x + 1}$, and $\mathrm{dv} = \frac{1}{x} ^ 2$ so $v = - \frac{1}{x}$

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

The latter integral can only be solved with parcial fractions, so we assume there is a sum of fractions

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

Where $a$ and $b$ are constants, that is also to say

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

Note that it means, that, for any value of $x$

$a \left(x + 1\right) + b x = 1$

So if we assume $x = 0$,

$a \left(0 + 1\right) + b \cdot 0 = 1$
$a = 1$

And if we assume $x = - 1$

$a \left(- 1 + 1\right) - b = 1$
$- b = 1$
$b = - 1$

So, back to the first integral we have

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

From there, it's an easy integral

$I = - \ln \frac{x + 1}{x} + \ln | x | - \ln | x + 1 | + c$