Integration by parts: $\int{\frac{dx}{(x^2 + a^2)^n}}$. I need to show that the following holds using integration by parts: 
\begin{equation}
\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}}}
\end{equation}
I really just don’t know where to start. It’s trivial to construct some solution of the form $\int u'v dx = uv - \int uv' dx$ to the integral on the left, but I can’t see how to get at this exact one. 
EDIT: 
I have tried to solve it by splitting it up, 
\begin{equation}
\int{\frac{dx}{(x^2 + a^2)^n}} = \int{\Big( \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{(x^2 + a^2)} \Big)}dx
\end{equation}
but as far as I can tell this results in something rather different from where I am supposed to end up: 
\begin{equation}
\int{\Big( \frac{1}{(x^2 + a^2)^{n-1}} \cdot \frac{1}{(x^2 + a^2)} \Big)}dx = \frac{1}{a}arctan \Big(\frac{x}{a}\Big) \cdot \frac{1}{(x^2 + a^2)^{n-1}} + \frac{2(n-1)}{a} \int{\frac{x^2 arctan \big(\frac{x}{a}\big)}{(x^2 + a^2)^{n}}}dx
\end{equation}
It is quite possible that I have made a very obvious mistake, so apologies in advance. 
I have also tried this: 
\begin{equation}
\int{(x^2 + a^2)^{-n}dx} = \int{\Big(1 \cdot (x^2 + a^2)^{-n}\Big) dx}
= x(x^2 + a^2)^{-n} + n\int{\frac{2x^2}{(x^2 + a^2)^{n+1}}  dx}
\end{equation}
Again, it doesn’t seem to lead me nearer the specific solution I need. 
 A: Let $I_{n}=\int \frac{dx}{(x^{2}+a^{2})^{n}}$, then
\begin{align*}
  I_{n} &=
  \frac{x}{(x^{2}+a^{2})^{n}}-
  \int x \, d\left[\frac{1}{(x^{2}+a^{2})^{n}} \right] \\
  &=\frac{x}{(x^{2}+a^{2})^{n}}+2n\int \frac{x^{2}dx}{(x^{2}+a^{2})^{n+1}} \\
  &=\frac{x}{(x^{2}+a^{2})^{n}}+
    2n\int \left[ \frac{1}{(x^{2}+a^{2})^{n}}-\frac{a^{2}}{(x^{2}+a^{2})^{n+1}}
           \right] dx \\
  &=\frac{x}{(x^{2}+a^{2})^{n}}+
    2nI_{n}-2a^{2}nI_{n+1} \\
  2a^{2}nI_{n+1} &=\frac{x}{(x^{2}+a^{2})^{n}}+(2n-1)I_{n} \\
  I_{n+1} &=
  \frac{x}{2a^{2}n(x^{2}+a^{2})^{n}}+\frac{2n-1}{2a^{2}n}I_{n} \\
  I_{n} &=
  \frac{x}{2a^{2}(n-1)(x^{2}+a^{2})^{n-1}}+\frac{2n-3}{2a^{2}(n-1)}I_{n-1} 
\end{align*}
A: Hint:
Use integration by parts with $\;\displaystyle \int\dfrac{\mathrm d\mkern1mu x}{(x^2+a^2)^{n-1}}$, setting
$$u=\frac1{(x^2+a^2)^{n-1}},\quad\mathrm d\mkern1mu v=\mathrm d\mkern1mu x.$$
A: Standart way of such:
$$J = \int\dfrac{dx}{(x^2+a^2)^n} = \dfrac1{a^2}\int\dfrac{(a^2+x^2)-x^2}{(x^2+a^2)^n}dx$$
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} - \dfrac1{a^2}\int\dfrac{x^2}{(x^2+a^2)^n}dx$$
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} + \dfrac{1}{2a^2(n-1)}\int x\,d\, \dfrac1{(x^2+a^2)^{n-1}}.$$
By parts:
$$J = \dfrac1{a^2}\int\dfrac{dx}{(x^2+a^2)^{n-1}} + \dfrac{1}{2a^2(n-1)} \dfrac {x}{(x^2+a^2)^{n-1}} - \dfrac{1}{2a^2(n-1)}\int \dfrac{dx}{(x^2+a^2)^{n-1}}, $$
$$\boxed{J = \dfrac{1}{2a^2(n-1)} \dfrac {x}{(x^2+a^2)^{n-1}} + \dfrac{2n-3}{2a^2(n-1)}\int \dfrac{dx}{(x^2+a^2)^{n-1}}}$$
