# Thread: Integration by parts/ reduction formula problem

1. ## Integration by parts/ reduction formula problem

Hi, not looking for an answer here, just the first step or two to get me started

Prove

$\int\frac{x^2}{(a^2+x^2)^n}\, dx= \frac{1}{2n-2}\left(-\frac{x}{(a^2+x^2)^{n-1}}+\int\frac{dx}{(a^2+x^2)^{n-1}}\right)$

Much appreciated

Stonehambey

2. That's not so complicated:
First note that:

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

That's a basic integral, and it easily follows from the following fundamental integral:

$
\int {x^n dx = \frac{1}
{{n + 1}}x^{n + 1} + C}
$

Now we can write the original integral as a multiple of two parts like so:

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

we already know the integral of the second part so by using integration by parts we get:

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

3. Thanks, it was that first step I didn't spot. As you said, it's pretty simple after that