# Thread: [SOLVED] Antiderivative

1. ## [SOLVED] Antiderivative

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

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

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

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

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

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

now backsubstitute:

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

simplify:

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

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

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