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Math Help - Integration by parts/ reduction formula problem

  1. #1
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    Nov 2008
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    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
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  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
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    That's not so complicated:
    First note that:

    <br />
\int {\frac{x}<br />
{{\left( {x^2  + a^2 } \right)^n }}dx =  - \frac{1}<br />
{{2\left( {n - 1} \right)}}\frac{1}<br />
{{\left( {x^2  + a^2 } \right)^{n - 1} }}} <br />

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

    <br />
\int {x^n dx = \frac{1}<br />
{{n + 1}}x^{n + 1}  + C} <br />

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

    <br />
\int {x\frac{x}<br />
{{\left( {x^2  + a^2 } \right)^n }}dx} <br />

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

    <br />
\int {x\frac{x}<br />
{{\left( {x^2  + a^2 } \right)^n }}dx}  =  - \frac{x}<br />
{{2\left( {n - 1} \right)}}\frac{1}<br />
{{\left( {x^2  + a^2 } \right)^{n - 1} }} + \frac{1}<br />
{{2\left( {n - 1} \right)}}\int {\frac{1}<br />
{{\left( {x^2  + a^2 } \right)^{n - 1} }}dx} <br />
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  3. #3
    Member
    Joined
    Nov 2008
    Posts
    114
    Thanks, it was that first step I didn't spot. As you said, it's pretty simple after that
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