Find $$I_n(a)=\displaystyle \int \limits_0^1 \dfrac{dx}{(x^2+a^2)^n}, \, a\ne0, \, 0\ne n \in \mathbb{N}.$$
I have following ideas. We have $I_n(a)=I_n(-a)$. \begin{align} \forall \,0\ne n \in \mathbb{N},&\, f(x,a)=\dfrac{1}{(x^2+a^2)^n}\, \text{is continuous on}\, D=[0,1]\times(0; \infty)\\ &f^{'}_a(x,a)=\dfrac{-2na}{(x^2+a^2)^{n+1}}\, \text{is continuous on}\, D. \end{align}
So, $$I^{'}_n(a)=\displaystyle \int \limits_0^1 \dfrac{-2nadx}{(x^2+a^2)^{n+1}}=-2naI_{n+1}(a)$$
I just think here.
Hint:
Traditionally, this kind of integral is computed recursively. The recurrence relation is obtained through an integration by parts: set $$\begin{cases} u=\dfrac1{(x^2+a^2)^n},\\ \mathrm dv=\mathrm d x, \end{cases}\quad\text{whence}\quad\begin{cases}\mathrm d u=-\dfrac{2nx\,\mathrm dx}{(x^2+a^2)^{n+1}},\\ v=x\end{cases}.$$ Integration by parts then yields $$I_n(a)=\frac x{(x^2+a^2)^n}\biggr\rvert_0^1+2n\underbrace{\int_0^1\frac{x^2\,\mathrm dx}{(x^2+a^2)^{n+1}}}_{I_n-a^2I_{n+1}}=\cdots$$
$$I_n(a)=\int_0^1\frac{dx}{(x^2+a^2)^n}$$ If: $$u=\frac{1}{(x^2+a^2)^n}\Rightarrow u'=\frac{-2nx}{(x^2+a^2)^{n+1}}$$ $$v'=1\Rightarrow v=x$$ so by IBP: $$I_n(a)=\left[\frac{x}{(x^2+a^2)^n}\right]_0^1+\int_0^1\frac{2nx^2}{(x^2+a^2)^{n+1}}$$ $$\Rightarrow I_n(a)=\frac{1}{(1+a^2)^n}+2n\int_0^1\left(\frac{1}{(x^2+a^2)^n}-\frac{a^2}{(x^2+a^2)^{n+1}}\right)dx$$ $$\Rightarrow I_n=\frac{1}{(1+a^2)^n}+2n\left(I_n-a^2I_{n+1}\right)$$ Now rearrange for $I_{n+1}$: $$2na^2I_{n+1}=\frac{1}{(1+a^2)^n}+(2n-1)I_n$$ And using this multiple times for a given $n$ will get you down to a final integral $I_1$ which is not too bad to work out :)
