Simple integration based on chain rule

[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

 

[radou]

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

 

[Office_Shredder]

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

 

[mbrmbrg]

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