Reduction formula for $\int\frac{dx}{(x^2+a^2)^n}$ How can I use integration by parts to write $$\int\frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}\int\frac{dx}{(x^2+a^2)^{n-1}}?$$
I would try $$u=\frac{1}{(x^2+a^2)^n},du=\frac{-2nx}{(x^2+a^2)^{n+1}}dx; dv=dx,v=x.$$ Integration by parts implies $$\int\frac{dx}{(x^2+a^2)^n}=\frac{x}{(x^2+a^2)^n}+2n\int\frac{x^2}{(x^2+a^2)^{n+1}}\\=\frac{x}{(x^2+a^2)^n}+2n\int\frac{(x^2+a^2)-a^2}{(x^2+a^2)^{n+1}}\\=\frac{x}{(x^2+a^2)^n}+2n\int\frac{1}{(x^2+a^2)^{n}}-2na^2\int\frac{dx}{(x^2+a^2)^{n+1}}$$ but I don't think I'm doing this correctly since the power of the denominator $(x^2+a^2)$ is not decreasing.
 A: Call your integral $I_n$ so that the result of your integration by parts is $$I_n=\frac{x}{(x^2+a^2)^n}+2nI_n-2na^2I_{n+1}$$
Now rearrange so that $I_{n+1}$ is the subject of the formula.
Then replace all $n$ with $n-1$ and you have the formula you require.
A: Hint:
Use integration by parts for the computation of $\;\displaystyle\int\frac{\mathrm dx}{(x^2+a^2)^{\color{red}{n-1}}} $, setting
$$ \begin{cases}u=\dfrac{1}{(x^2+a^2)^{n-1}},\\[1ex]\mathrm dv=\mathrm dx,\end{cases}
\;\text{ whence}\quad\begin{cases}\mathrm du=\dfrac{-2(n-1)x\,\mathrm dx}{(x^2+a^2)^{n}},\\[1ex] v=x,\end{cases} $$
so that
\begin{align}
\int\frac{\mathrm dx}{(x^2+a^2)^{n-1}}&=\dfrac{x}{(x^2+a^2)^{n-1}}+2(n-1)\int \dfrac{x^2\,\mathrm dx}{(x^2+a^2)^{n}} \\
&=\dfrac{x}{(x^2+a^2)^{n-1}}+2(n-1)\int \dfrac{\mathrm dx}{(x^2+a^2)^{n-1}}-2(n-1)a^2\int \dfrac{\mathrm dx}{(x^2+a^2)^{n}}\\\text{and finally}\hspace4em\\[1ex]
2(n-1)a^2\int \dfrac{\mathrm dx}{(x^2+a^2)^{n}}&=\dfrac{x}{(x^2+a^2)^{n-1}}+(2n-3)\int \dfrac{\mathrm dx}{(x^2+a^2)^{n-1}}.
\end{align}
A: hint
$$\frac {1}{(x^2+a^2)^n}=\frac {1}{a^2}\frac {a^2+x^2-\frac {2x}{2}x}{(a^2+x^2)^n} $$
A: Let $a\neq0$, $n\in\mathbb{N}$ and define, $$I_n:=\int{\frac{dx}{(x^2+a^2)^n}}$$
So, If we use integral by parts, we have $$I_n = \frac{x}{(x^2+a^2)^n}+2n\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}\qquad(*)$$
On the other hand, is clear that $$I_n=\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}+a^2\int{\frac{dx}{(x^2+a^2)^{n+1}}}=\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}+a^2I_{n+1}\qquad(**)$$
Now using $(*)$ and $(**)$, follows that $$I_n=\frac{x^2}{(x^2+a^2)^{n}}+2n(I_n-a^2I_{n+1})\implies I_{n+1}=\frac{1}{2a^{2}n}\left(\frac{x}{(x^2+a^2)^n}-(1-2n)I_n\right)$$
In particular, if we replace $n$ by $n-1$, we have 
$$I_n=\frac{1}{2a^{2}(n-1)}\left(\frac{x}{(x^2+a^2)^{n-1}}-(3-2n)I_{n-1}\right)$$
