How to evaluate $\int \sqrt{x^3+x^4} dx$

Question

Find the integral: $\int \sqrt{x^3+x^4} dx$

I know I could use WolframAlpha, but I wonder if there is a way to calculate this integral in a nice way (some clever substitution perhaps?).

Tried a lot of different ways, but don't seem to find a good way.

I don't want you to calculate the whole integral, I just ask you for a little hint.

I know. I just hoped to find one a little prettier than the Wolfram's. — Mateusz, May 06 '14 at 19:29
@user142866 Hope is a good thing. Maybe, the best of things. — The very fluffy Panda, May 06 '14 at 19:31
A pity: if it were $;x^2;$ instead of that nasty $;x^3;$ then there is a slick solution... — DonAntonio, May 06 '14 at 19:32
@DonAntonio: This problem is tagged as "hard" in my coursebook. But you're right, if it were $x^2$ I wouldn't need to seek for some help :) — Mateusz, May 06 '14 at 19:35

score 4 · Accepted Answer · answered May 06 '14 at 19:42

Here is a suggested method, which involves completing the square and then two substitutions.

$$\sqrt{x^3+x^4}=\frac x2\sqrt{(2x+1)^2-1}$$

Now set $y=2x+1, dy=2dx, x=\frac {y-1}2$ and the integrand becomes $$\frac {y-1}8\cdot\sqrt {y^2-1}$$

Then use the substitution $y=\cosh z$

1 Answers1