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The question is to "prove the reduction formula"

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

What I got is


$ u = x $

$ du = dx $

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

$\displaystyle{ v = \frac{ 1 }{ 2(n+1) \left( a^2 + x^2 \right)^{n+1} } }$

So I got

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

Which I believe is correct. They are subtracting from n in the integration step and I'm not sure why

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+1 for, as usual, showing your work. – Arturo Magidin Apr 3 '11 at 1:19
up vote 6 down vote accepted

You went wrong when you integrated $dv$.

You have $dv = x(a^2+x^2)^{-n}\,dx$. When you integrate, you add one to the exponent. But adding one to $-n$ gives $-n+1 = -(n-1)$. So $$v = \frac{1}{2(-n+1)}(a^2+x^2)^{-n+1} = \frac{1}{2(1-n)(a^2+x^2)^{n-1}}.$$ The minus sign from integration by parts can be cancelled out by switching the sign of $2(1-n)$ to get $2(n-1) = 2n-2$.

If you use the correct value of $v$, I think you will have no trouble establishing the formula.

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+1 for the first sentence. – awllower Apr 3 '11 at 4:17

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