Integration of functions of type $\sqrt[m]{k\alpha x^2}$

Question

I would like some help in integrating a function like that:

$$\int\sqrt[3]{4\alpha x^2}dx$$ for $a\in \mathbb{R}^+.$

More generally, what's the heuristic for dealing with functions of the type:

$$\int\sqrt[m]{k\alpha x^2}dx$$ for $m\in \mathbb{N}^*,$ $k\in \mathbb{N}^*,$ $a\in \mathbb{R}^+.$

What about $$\sqrt[m]{kP(x)}dx$$ for $m\in\mathbb{N}^*, \ k\in \mathbb{N}^*,$ $a\in \mathbb{R}^+ $ and $P(x)$ a polynomial (you can add more hypotheses if needed) ?

Welcome to Mathematics Stack Exchange. $\large\int\sqrt[3]{4\alpha x^2}dx=$ $(4\alpha)^{1/3}\large\int x^{2/3}dx$ $=(4\alpha)^{1/3}\dfrac35 x^{(5/3)}+C$ — J. W. Tanner, Feb 17 '20 at 04:31
Is there anything you can tell us about the function $P(x)$? Is $P(x)$ a polynomial? — user729424, Feb 17 '20 at 04:34
Integral $\int\sqrt{P(x)};dx$ where $P$ is a polynomial of degree $3$ or $4$ is an elliptic integral. — GEdgar, Feb 18 '20 at 00:46

score 2 · Accepted Answer · answered Feb 17 '20 at 04:33

2

Well the first integral is really easy. You just need to separate the constants out. $$I=\int\sqrt[m]{k\alpha x^2}\,dx$$ $$I=\sqrt[m]{k\alpha}\int x^{\frac2m}\,dx$$ $$I=\sqrt[m]{k\alpha}\cdot\frac {x^{\frac 2m+1}}{\frac 2m+1}$$ As for the second part, the method of solving depends highly on the nature of $P(x)$. There is no general solution to the integrals in this form.

answered Feb 17 '20 at 04:33

Sam

2,447

+1. I'm waiting to see if someone adds anything useful about the general polynomial case before accepting – Feb 17 '20 at 04:40
A similar question here shows general solutions for $m=2$ – Sam Feb 17 '20 at 04:43

Integration of functions of type $\sqrt[m]{k\alpha x^2}$

1 Answers1