1. ## [SOLVED] Antiderivative

I don't really know how to approach this one:

$\int\frac{dx}{x(1+x^n)}$

2. $\int {\frac{1}
{{x\left( {1 + x^n } \right)}}dx}
$

$\begin{gathered}
x = \left( {\tan u} \right)^{\frac{2}
{n}} \hfill \\
dx = \frac{2}
{n}\left( {\tan u} \right)^{\frac{2}
{n} - 1} \frac{1}
{{\cos ^2 u}} \hfill \\
\end{gathered}$

$
\int {\frac{{\frac{2}
{n}\left( {\tan u} \right)^{\frac{2}
{n} - 1} \frac{1}
{{\cos ^2 u}}}}
{{\left( {\tan u} \right)^{\frac{2}
{n}} \left( {1 + \tan ^2 u} \right)}}} du
$

$
\int {\frac{2}
{n}\cot udu = \frac{2}
{n}\ln \left| {\sin u} \right| + C}
$

now backsubstitute:

$
{\frac{2}
{n}\ln \left| {\sin a\tan x^{\frac{n}
{2}} } \right| + C}
$

simplify:

$
\frac{1}
{n}\ln \left| {\frac{{x^n }}
{{1 + x^n }}} \right| + C
$

3. Originally Posted by Spec
$\int\frac{dx}{x(1+x^n)}$
$\int {\frac{1}
{{x\left( {1 + x^n } \right)}}\,dx} = \int {\frac{{\left( {1 + x^n } \right) - x^n }}
{{x\left( {1 + x^n } \right)}}\,dx} = \int {\frac{1}
{x}\,dx} - \frac{1}
{n}\int {\frac{{\left( {1 + x^n } \right)'}}
{{1 + x^n }}\,dx} .$

Hence $\int {\frac{1}
{{x\left( {1 + x^n } \right)}}\,dx} = \ln \left| x \right| - \frac{1}
{n}\ln \left( {1 + x^n } \right) + k.$