I have to prove this formula:
$$\int \! \frac{1}{(x^2+\beta ^2)^{k+1}} \, \mathrm{d}x=\frac{1}{2k\beta ^2}\frac{x}{(x^2 +\beta^2)^k}+\frac{2k-1}{2k\beta^2}\int \! \frac{1}{(x^2+\beta ^2)^k} \, \mathrm{d}x $$
I have to use partial fraction decomposition. I started like this: $$ \frac{a}{(x^2+\beta ^2)^k}+\frac{b}{x^2+\beta ^2} $$
Now: $$ 1 = a(x^2+\beta^2) + b(x^2+\beta^2)^k $$
Now I'm stuck. How do I find $a$ and $b$?
Write
$$\begin{align} \frac{1}{(x^2+\beta^2)^{k+1}}&=\frac{1}{\beta^2}\frac{(x^2+\beta^2)-x^2}{(x^2+\beta^2)^{k+1}}\\\\ &=\frac{1}{\beta^2}\left(\frac{1}{(x^2+\beta^2)^{k}}- \frac{x^2}{(x^2+\beta^2)^{k+1}}\right) \end{align}$$
Now, integrate both sides and use integration by parts on the last term. Combine terms of like powers and you will have it.
NOTE: Integration by parts
$$-\int \frac{x^2}{(x^2+\beta^2)^{k+1}} dx = \frac{1}{2k}\frac{x}{(x^2+\beta^2)^k}-\frac{1}{2k}\int \frac{1}{(x^2+\beta^2)^{k+1}}dx$$
