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?


2 Answers 2


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

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

Can you finish it?


Let us denote the integral by


Before we move on let us make the observations that




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


which we can rearrange to get


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


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.


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