# Evaluating $\int\limits_0^1 {\frac{{{x^{2n - 3}}}}{{{{(1 + {x^2})}^n}}}dx}$

Evaluate $$\displaystyle I=\int\limits_0^1 {\frac{{{x^{2n - 3}}}}{{{{(1 + {x^2})}^n}}}dx}$$

I tried substituting $$u=x^2+1$$ and $$du=2x$$, the bounds change from 1 to 2

Hence, $$I=\displaystyle\frac{1}{2}\int\limits_1^2 {\frac{{{{(u - 1)}^{n - 2}}}}{{{u^n}}}du}$$

This integral is really challenging to me, I have no idea how to get further.

Are there any better way than this?

$$I=\int\limits_0^1 {\frac{{{x^{2n - 3}}}}{{{{(1 + {x^2})}^n}}}dx}$$

$$I=\int\limits_0^1 {\frac{1}{x^3{{{(1 + {\frac{1}{x^2}})}^n}}}dx}$$

Can you finish it?

Let us denote the integral by

$$I_n=\int_0^1\frac{x^{2n-3}}{(1+x^2)^n}\mathrm{d}x.$$

Before we move on let us make the observations that

$$I_{n+1}=\int_0^1\frac{x^{2n-1}}{(1+x^2)^{n+1}}\mathrm{d}x$$

and

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{(1+x^2)^n}\right)=-\frac{2xn}{(1+x^2)^{n+1}}.$$

This gives us a small hint that maybe integration by parts will help us out somehow, so let us give it a try. Integration by parts (where we differentiate what I already differentiated and integrate what is left) gives us that

\begin{align*} I_n &= \int_0^1\frac{x^{2n-3}}{(1+x^2)^n}\mathrm{d}x \\ &= \left[\frac{x^{2n-2}}{2n-2}\cdot\frac{1}{(1+x^2)^n}\right]_0^1+\int_0^1\frac{x^{2n-2}}{2n-2}\cdot\frac{2xn}{(1+x^2)^{n+1}}\mathrm{d}x \\ &= \frac{1}{2^{n+1}(n-1)}+\frac{n}{n-1}\int_0^1\frac{x^{2n-1}}{(1+x^2)^{n+1}}\mathrm{d}x. \end{align*}

Now notice that something great happened: integration by parts gave us back something familiar! Indeed what we have now is that

$$I_n=\frac{1}{2^{n+1}(n-1)}+\frac{n}{n-1}I_{n+1},$$

which we can rearrange to get

$$I_{n+1}=\frac{n-1}{n}I_n-\frac{1}{2^{n+1}n}.$$

There is a subtilty in getting here to keep in mind though. If $$n=1$$, then the antiderivative of $$x^{2n-3}$$ is not $$x^{2n-2}$$, so this would only give us the solutions when $$n\geq 2$$ (you can probably spot that $$n=1$$ also gives problems in the original recursive formula we found). To finish this part off notice that, using the substitution you tried before, we can find that

$$I_2=\int_0^1\frac{x}{(1+x^2)^2}\mathrm{d}x=\frac{1}{4}.$$

Thus we have that the integral is recursively given by

$$\begin{cases}\displaystyle I_{n+1}=\frac{n-1}{n}I_n-\frac{1}{2^{n+1}n}, & n\geq 2, \\ I_2 = \frac{1}{4} \end{cases}.$$

This of course does not give you a closed form, but hopefully it gives you some valuable insight into the problem and how you can tackle problems like this, and if you really want a closed form, try to solve the recursion formula and see if it gets you anywhere.