Reduction formula proof

Question

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

Answer 1

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:

$$I_m=\frac{1}{a^2}\left(\frac{2m-3}{2m-2}I_{m-1}+\frac{u}{(2m-2)(a^2+u^2)^{m-1}}\right)$$

Answer 2

Alternatively, if you want to compute the integral $$\int_{a}^{b}\frac{3x-2}{(x^{2}+2x+2)^{2}}dx$$ directly, you can split it up. Indefinite integral is used as the bounds just get in the way.

$$\int\frac{3x-2}{(x^{2}+2x+2)^{2}}dx=\int\frac{3x+3dx}{(x^{2}+2x+2)^{2}} - 5\int \frac{dx}{(x^{2}+2x+2)^{2}}$$

The first integral is $$-\frac{3}{2}\cdot\frac{1}{x^{2}+2x+2}$$ by the chain rule. Letting $u=x+1$, the, we have $$\begin{align*} \int \frac{dx}{(x^{2}+2x+2)^{2}} & = \int\frac{du}{(u^{2}+1)^{2}}\\ & = \frac{1}{2}\int \frac{1+u^{2}}{(1+u^{2})^{2}}+\frac{1-u^{2}}{(1+u^{2})^{2}}du\\ & = \frac{1}{2}\arctan(u) + \frac{1}{2}\int \frac{(1+u^{2})\cdot 1 - u\cdot 2u}{(1+u^{2})^{2}}du\\ & = \frac{1}{2}\arctan(u)+\frac{1}{2}\cdot\frac{u}{(1+u^{2})^{2}}\qquad\text{by the quotient rule} \end{align*}$$

Now just substitute $u=x+1$ back in and you have your integral.