The integral in which I am interested in is $$\int x(x^3+1)^{33}\mathrm{d}x$$

I tried to solve by substituting $x^2 = t$, but it didn't help. I find a solution by expanding it with the help of binomial expansion. Can anyone help me with any other method like substitution, by parts?

  • $\begingroup$ Repeated IBP 33 times leads to $\int x^{100}dx$ $\endgroup$ – Quanto Aug 13 at 21:05
  • $\begingroup$ I can't get it!! can you explain in detail please?? $\endgroup$ – user579689 Aug 13 at 21:09
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    $\begingroup$ Given how terribly long the answers here are, I suspect that there was a typo in the original problem $\endgroup$ – Omnomnomnom Aug 13 at 21:22
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    $\begingroup$ To elaborate @Omnomnomnom's comment, if the integral instead were $\int x^\color{red}{2} (x^3 + 1)^{33} \textrm{d}x$, the integral would be manageable with a straightforward substitution---and failing to see that substitution would force one to use one of the less pleasant techniques apparently necessary here. $\endgroup$ – Travis Aug 13 at 22:17
  • $\begingroup$ @Quanto: The Binomial Theorem gives a most significant term of $\int x^{100}\,\mathrm{d}x$. That might be the best way to attack this. $\endgroup$ – robjohn Aug 13 at 23:35

This is not easier than expanding using the Binomial theorem, but it's a different way to approach it which you may at least find interesting, and even potentially useful (in other situations if not this one).

For any integer $n \ge 0$, let

$$f(n) = \int x(x^3 + 1)^n dx \tag{1}\label{eq1}$$

For $n \ge 1$, using integration by parts, where $u(x) = (x^3 + 1)^n$ so $d(u(x)) = 3nx^2(x^3 + 1)^{n-1}dx$, and $d(v(x)) = xdx$ so $v(x) = \frac{x^2}{2}$, you get

$$\begin{equation}\begin{aligned} f(n) & = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \int x^4(x^3 + 1)^{n-1} dx \\ & = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \int x(x^3 + 1 - 1)(x^3 + 1)^{n-1} dx \\ & = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \int \left(x(x^3 + 1)^n - x(x^3 + 1)^{n-1}\right) dx \\ & = \frac{x^2}{2}(x^3 + 1)^n - \frac{3n}{2} \left(f(n) - f(n-1)\right) \end{aligned}\end{equation}\tag{2}\label{eq2}$$

This leads to the recursive equation

$$\begin{equation}\begin{aligned} \left(1 + \frac{3n}{2}\right)f(n) & = \frac{x^2}{2}(x^3 + 1)^n + \frac{3n}{2}f(n-1) \\ f(n) & = \frac{x^2}{2 + 3n}(x^3 + 1)^n + \frac{3n}{3n + 2}f(n-1) \end{aligned}\end{equation}\tag{3}\label{eq3}$$

You can determine what $f(0)$ is (I'm leaving that to you) and then use \eqref{eq3} to determine each of the rest of the $f$ values up to $f(33)$.


Repeat the integral-by-parts 33 times as follows:

$$I_0=\frac{1}{2}\int (x^3+1)^{33} dx^2 = \frac{1}{2}x^2 (x^3+1)^{33} -\frac{99}{2}I_1$$

$$I_1= \int x^4 (x^3+1)^{32} dx = \frac{1}{5}x^5 (x^3+1)^{32 } -\frac{96}{5}I_2 $$

$$I_2=\int x^7 (x^3+1)^{31} dx = \space ... $$


$$I_{32}=\int x^{97} (x^3+1) dx =\frac{1}{98}x^{98} (x^3+1) -\frac{3}{98}I_{33} $$

$$I_{33} = \int x^{100} dx$$

You should get the same result and the number of terms from your binomial expansion method.


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