I am looking to compute something like this : $\displaystyle\int\limits_a^b \dfrac{3x-2}{(x^2+2x+2)^2} dx$
Making some researches, it seems that an appropriate varaible change combined with completing the square will do the trick if I use the reduction formula $$ \displaystyle\int\dfrac{du}{(a^2+u)^m}=\frac1{a^2}\left(\dfrac{u}{\left(2m-2\right)\left(a^2+u^2\right)^{m-1}}+\dfrac{2m-3}{2m-2}\int\dfrac{du}{\left(a^2+u^2\right)^{m-1}}\right), m\neq 1$$
formula I found in a book of mine.
I can compute the integral using this reduction formula, my question is how you prove it ? I couldn't manage it, I guess there it has to be part integrated, but all my attempts failed badly.
Thanks in advance
Best Answer
Actually, you can prove that this formula is correct just by taking derivatives of both sides of the equation.
Otherwise, if you want to derive this formula you can apply integration by parts:
$$I_m=\int\dfrac{du}{(a^2+u^2)^m}=\frac{u}{(a^2+u^2)^m}-\int ud\frac{1}{(a^2+u^2)^m}=\frac{u}{(a^2+u^2)^m}+2m\int \frac{u^2du}{(a^2+u^2)^{m+1}}=\frac{u}{(a^2+u^2)^m}+2m\int \frac{((u^2+a^2)-a^2)du}{(a^2+u^2)^{m+1}}=\frac{u}{(a^2+u^2)^m}+2mI_m-2ma^2I_{m+1}$$
So: $I_{m+1}=\frac{1}{2ma^2}((2m-1)I_m+\frac{u}{(a^2+u^2)^m})$ or:
$$I_m=\frac{1}{a^2}\left(\frac{2m-3}{2m-2}I_{m-1}+\frac{u}{(2m-2)(a^2+u^2)^{m-1}}\right)$$