Prove the reduction formula 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 
Set
$ 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

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.
