# Calculus - indefinite integration

by user579689   Last Updated August 13, 2019 21:20 PM

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?

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Repeat the integral-by-parts below 33 times:

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

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

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

$$...$$

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

Quanto
August 13, 2019 21:17 PM

This is not necessarily any easier than expanding using the Binomial theorem, but it's a different way to approach it which you may find useful.

For any $$n \in \mathbb{N}$$, let

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

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 and then use \eqref{eq3} to determine each of the rest of the $$f$$ values up to $$f(33)$$.

John Omielan
August 13, 2019 21:19 PM