# Prove that $\int{\frac{x^2}{(x^2 + a^2)^n}}dx = \int{\frac{1}{(x^2 + a^2)^n}}dx - a^2\int{\frac{1}{(x^2 + a^2)^{n+1}}}dx$

I am trying to derive the following reduction formula for integrating $\int{\frac{1}{(x^2 + a^2)^n}}dx, n,a>0$: $$I_n = \frac{1}{2a^2(n-1)}\left(\frac{x}{(x^2 + a^2)^{n-1}} + (2n-3)I_{n-1} \right)$$ So far I got $$\int{\frac{1}{(x^2 + a^2)^n}}dx = \frac{x}{(x^2 + a^2)^n} + 2n\int{\frac{x^2}{(x^2 + a^2)^{n+1}}}dx$$ and $$I_n = \frac{x}{(x^2 + a^2)^n} + 2nI_n - 2na^2I_{n+1} \Rightarrow I_n = \frac{1}{2a^2(n-1)}\left(\frac{x}{(x^2 + a^2)^{n-1}} + (2n-3)I_{n-1} \right)$$ but I'm struggling with the step in between: $$\int{\frac{x^2}{(x^2 + a^2)^n}}dx = I_n - a^2I_{n+1}$$

• You have an ordinary identity: $\frac{x^2}{(x^2 + a^2)^{n+1}}= \frac{1}{(x^2 + a^2)^n} - a^2\frac{1}{(x^2 + a^2)^{n+1}}$... Are you sure the exponent in the first integral is $n$ and not $n+1$? – user491874 Feb 12 '18 at 20:25
• Also, $n=0$: $\int{x^2dx}=\int{dx}-a^2\int{\frac{1}{x^2+a^2}dx}$ doesn't seem likely to hold, e.g. for $a=1$ you would have $\frac{x^3}{3}=x-\arctan x+C$... – user491874 Feb 12 '18 at 20:28
• @user8734617 it holds for n>0, I added a clarification, thank you. – ZyTelevan Feb 12 '18 at 20:31
• That equation doesn't hold for $a=0$ – Jose Lopez Garcia Feb 12 '18 at 20:34

$$\frac{1}{(x^2+a^2)^{n}}-\frac{a^2}{(x^2+a^2)^{n+1}}=$$ $$\frac{x^2+a^2}{(x^2+a^2)^{n+1}}-\frac{a^2}{(x^2+a^2)^{n+1}}=$$ $$\frac{x^2+a^2-a^2}{(x^2+a^2)^{n+1}}=$$ $$\frac{x^2}{(x^2+a^2)^{n+1}}\neq\frac{x^2}{(x^2+a^2)^{n}}$$

• Oh. I tried integrating by parts again and u-substitution, but missed something that elementary, thank you! – ZyTelevan Feb 12 '18 at 20:35
• @ZyTelevan You're welcome. – Botond Feb 12 '18 at 20:36

Let $$a\neq0$$, $$n\in\mathbb{N}$$ and define, $$I_n:=\int{\frac{dx}{(x^2+a^2)^n}}$$

So, If we use integral by parts, we have $$I_n = \frac{x}{(x^2+a^2)^n}+2n\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}\qquad(*)$$

On the other hand, $$I_n=\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}+a^2\int{\frac{dx}{(x^2+a^2)^{n+1}}}=\int{\frac{x^2}{(x^2+a^2)^{n+1}}dx}+a^2I_{n+1}\qquad(**)$$

Now using $$(*)$$ and $$(**)$$, follows that $$I_n=\frac{x^2}{(x^2+a^2)^{n}}+2n(I_n-a^2I_{n+1})\implies I_{n+1}=\frac{1}{2a^{2}n}\left(\frac{x}{(x^2+a^2)^n}-(1-2n)I_n\right)$$

In particular, if we take $$n-1$$, we have

$$I_n=\frac{1}{2a^{2}(n-1)}\left(\frac{x}{(x^2+a^2)^{n-1}}-(3-2n)I_{n-1}\right)$$