Could someone please show me how to do the indefinite integral of
$$\frac{1}{(ax^2+bx+c)^n}$$
a) using real analysis (hard)
b) using complex analysis (nice factoring)
and show they give the same answer, without using any simplifiers into $1 + t^2$ or other stuff making it easier (but assumes you remember the answer or what to do).
I keep getting incorrect answers and forgetting what to do :(
$$\int_{0}^{+\infty}\frac{dt}{(1+t^2)^n}=\int_{0}^{\pi/2}\cos^{2n-2}(\theta)\,d\theta = \frac{2\pi}{4^{n}}\cdot\binom{2n-2}{n-1},\tag{1}$$ where the first identity depends on the substitution $t=\tan\theta$ and the second one follows by integrating by parts. In the parametric case: $$\begin{eqnarray*}\int_{-\infty}^{+\infty}\frac{dx}{(ax^2+bx+c)^n}&=&(4a)^n\int_{-\infty}^{+\infty}\frac{dy}{((2ay+b)^2+(4ac-b^2))^n}\\ &=&(4a)^n\int_{-\infty}^{+\infty}\frac{dz}{((2az)^2+(4ac-b^2))^n}\\ &=& 2^{2n-1}a^{n-1}\int_{-\infty}^{+\infty}\frac{dz}{(z^2+(4ac-b^2))^n}\\ &=& 2^{2n-1}a^{n-1}(4ac-b^2)^{1/2-n}\int_{-\infty}^{+\infty}\frac{dw}{(w^2+1)^n}\\ &=& 2\pi\cdot a^{n-1}(4ac-b^2)^{1/2-n}\cdot\binom{2n-2}{n-1}.\\ \end{eqnarray*}$$ Indefinite integration can be carried over through the following observation: it is easy to integrate $\cos^{2n}(\theta)$ because from the De Moivre identity you know its Fourier series: $$\cos^{2n}(\theta)=\frac{1}{4^n}\sum_{k=0}^{2n}\binom{2n}{k}e^{i(2n-2k)\theta}=\frac{2}{4^n}\left(\binom{2n}{n}+2\sum_{k=0}^{n-1}\binom{2n}{k}\cos((2n-2k)\theta)\right),$$ $$\int\cos^{2n}(\theta)d\theta = \frac{2}{4^n}\left(\binom{2n}{n}\theta+2\sum_{k=0}^{n-1}\binom{2n}{k}\frac{\sin((2n-2k)\theta)}{2n-2k}\right).$$ To recover the indefinite integral as a function of $t$, it suffices to expand the Chebyshev polynomials: $$\sin((2n-k)\theta) = \sin(\theta)\cdot U_{2n-k-1}(\cos\theta) = \frac{t}{\sqrt{1+t^2}}U_{2n-k-1}\left(\frac{1}{\sqrt{1+t^2}}\right).$$ As an alternative, consider that integration by parts gives: $$ I_n=\int \frac{dt}{(1+t^2)^n} = I_{n-1}-\int t\cdot\frac{t}{(1+t^2)^n} = I_{n-1}+\frac{t}{(2n-2)(1+t^2)^{n-1}}-\frac{I_n}{2n-2},$$ $$I_{n}=\frac{2n-3}{2n-2}I_{n-1}+\frac{1}{2n-2}\cdot\frac{t}{(1+t^2)^{n-1}},$$ hence: