# Simple integration based on chain rule

1. Oct 10, 2006

### mbrmbrg

I think I've got this, but I'm not quite sure, especially about multiplying to get du to be what I want. Can someone please tell me if this is correct? (The first line is the problem.) Thanks!

$$\int\frac{3xdx}{\sqrt[3]{2x^2+3}}$$
I set $$u=2x^2+3, du=4x$$ so the problem becomes
$$\int u^{-\frac{1}{3}}(\frac{3}{4})(\frac{4}{3})3xdx$$
rearranging the constants gives the much nicer equation
$$\frac{3}{4}\int u^{-\frac{1}{3}}du$$
$$=(\frac{3}{4})(\frac{u^\frac{2}{3}}{\frac{2}{3}})+C$$
$$=\frac{9(2x^2+3)^\frac{2}{3}}{8}+C$$

Last edited: Oct 10, 2006
2. Oct 10, 2006

Correct. Just one detail, though: du = 4xdx, but you probably just let it out by mistake, since the solution is correct.

3. Oct 10, 2006

### Office_Shredder

Staff Emeritus
The best way to check your work is to just take the derivative, and see if it matches what you integrated in the first place

4. Oct 10, 2006

### mbrmbrg

Thank you very much!
I can now happily continue my four sections of calc homework that I wisely saved for long lazy days...