Reduction formula for $$I_n =\int {x^{n}\over \sqrt{x^2 + a}} dx$$
$\displaystyle u(x) = {x^{n}\over \sqrt{x^2 + a}}, \qquad\qquad v^{'}(x) = 1$
Integrating by parts I get,
$$I_n = {x^{n+1}\over \sqrt{x^2 + a}} - nI_n + \int {x^{n+2}\over (x^2 + a)^{3/2}}dx$$
The problem is $J =\displaystyle\int {x^{n+2}\over (x^2 + a)^{3/2}}dx$, I tried to partial fraction decomposition on ${x^{n+2}\over (x^2 + a)^{3/2}}$ but the result is very messy.
How to write $J$ in terms of $I$ ?
There are probably simpler ways, but let me give one.
Considering $$I_n=\int {x^{n}\over \sqrt{x^2 + a}} dx$$ change variable $$x=\sqrt a \sinh(y)\implies dx=\sqrt a \cosh(y)\,dy$$ which make $$I_n=a^{n/2}\int\sinh ^n(y)\,dy$$ Now, let $y=iz$ which make $$I_n=i^{n+1}a^{n/2}\int \sin^n(z)\,dz$$ So $$J_n=\frac{I_n}{i^{n+1}a^{n/2}}=\int \sin^n(z)\,dz$$ for which we know the recursion formula (have a look here).
Integrate by parts \begin{align} I_n &= \int {x^{n}\over \sqrt{x^2 + a}} dx= \int x^{n-1}\ d\left( \sqrt{x^2+a}\right)\\ &=\ x^{n-1} \sqrt{x^2+a}-(n-1) \int x^{n-2}\frac{x^2+a}{\sqrt{x^2+a}}dx\\ &=\ x^{n-1} \sqrt{x^2+a}-(n-1)I_n -(n-1)aI_{n-2} \end{align} which yields $$I_n = \frac{x^{n-1} }n\sqrt{x^2+a}-\frac{(n-1)a}nI_{n-2} $$
Hint: Take $a=b^2$ and first of all get rid of $b$. Then do integration by parts of what left to get a recursion formula.
