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

Jun 3, 2018

$- \ln \frac{x + 1}{x} + \ln | \frac{x}{x + 1} | + C , \mathmr{and} ,$

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

#### Explanation:

The Rule of Integration by Parts :

$\int u v ' \mathrm{dx} = u v - \int u ' v \mathrm{dx}$.

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

We take, $u = \ln \left(x + 1\right) \mathmr{and} v ' = \frac{1}{x} ^ 2$.

$\therefore u ' = \frac{1}{x + 1} \mathmr{and} v = \int v ' \mathrm{dx} = \int \frac{1}{x} ^ 2 \mathrm{dx} = - \frac{1}{x}$.

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

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

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

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

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

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

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

$\Rightarrow I = - \ln \frac{x + 1}{x} + \ln | \frac{x}{x + 1} | + C ,$

$\mathmr{and} , I = - \frac{1}{x} \ln | x + 1 | + \ln | \frac{x}{x + 1} |$,

$= \ln {\left(x + 1\right)}^{- \frac{1}{x}} + \ln | \frac{x}{x + 1} |$,

$= \ln \left\{| {\left(x + 1\right)}^{- \frac{1}{x}} \cdot \frac{x}{x + 1} |\right\}$.

$\therefore I = \ln \left\{| x \cdot {\left(x + 1\right)}^{- \left(\frac{1}{x} + 1\right)} |\right\} + C$.

Enjoy Maths.!

Jun 3, 2018

$- \log \frac{x + 1}{x} + \log | x | - \log | x + 1 | + C$

#### Explanation:

We use Integration by parts:
$\int f \mathrm{dg} = f g - \int g \mathrm{df}$

with
$f = \log \left(x + 1\right) , \mathrm{dg} = \frac{1}{x} ^ 2 \mathrm{dx}$
we get
$\mathrm{df} = \frac{1}{x + 1} \mathrm{dx} , g = - \frac{1}{x}$
so we have
$- \log \frac{x + 1}{x} + \int \frac{1}{x \cdot \left(x + 1\right)} \mathrm{dx}$
$= \log \frac{x + 1}{x} + \int \frac{1}{x} \mathrm{dx} - \int \frac{1}{x + 1} \mathrm{dx}$
so we get
$- \log \frac{x + 1}{x} + \log | x | - \log | x + 1 | + C$