This is a very interesting problem that I came across in an old textbook of mine. So I know its got something to do with integrals, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

For every integer $m\geq 0$ let

$$I_m=\int_0^1x^m\left(x^2 -1 \right)^5dx$$

Prove that for $m\geq 2$

$$I_m= \frac{m-1}{m+11}\,I_{m-2}.$$

  • 1
    $\begingroup$ Write $x^m(x^2-1)^5=x^{m-1}x(x^2-1)^5$ and integrate by parts. $\endgroup$
    – Mark Viola
    May 28, 2016 at 4:12

1 Answer 1


Let's do the job as suggested by @Dr. MV. Integrate by parts with $u=x^{m-1}$ and $dv=x\left(x^2-1\right)^{5}dx$. This means $du=(m-1)x^{m-2}$ and $v={\left(x^2-1\right)^6\over 12}$. And so

$$\begin{align}I_m=&=\left[uv\right]_0^1-\int_0^1vdu\\ &=-{m-1\over 12}\int_0^1\left(x^2-1\right)^6x^{m-2}dx\\ &=-{m-1\over 12}\int_0^1\left(x^2-1\right)^5\left(x^2-1\right)x^{m-2}dx\\ &=-{m-1\over 12}\left(\int_0^1\left(x^2-1\right)^5x^{m}dx-\int_0^1\left(x^2-1\right)^5x^{m- 2}dx\right) \end{align}$$

This rewrites as follows

$$I_m=-{m-1\over 12}\left(I_m-I_{m-2}\right)$$

This gives by solving for $I_m$

$$I_m={m-1\over m+11}I_{m-2}$$

As expected

  • $\begingroup$ Well done! Very clear explanation. $\endgroup$
    – anonymous
    May 28, 2016 at 6:30

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