# Calculus - indefinite integration

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?

• Repeated IBP 33 times leads to $\int x^{100}dx$ – Quanto Aug 13 at 21:05
• I can't get it!! can you explain in detail please?? – user579689 Aug 13 at 21:09
• Given how terribly long the answers here are, I suspect that there was a typo in the original problem – Omnomnomnom Aug 13 at 21:22
• 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. – Travis Aug 13 at 22:17
• @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. – 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{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}\tag{2}\label{eq2}

This leads to the recursive 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}\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.