## TomLikesPhysics 2 years ago I need some help with an Integral (Substitution).

1. TomLikesPhysics

$\int\limits_{?}^{?}x^2*\sqrt{3x^2+4}dx$ I started with $u = 2x^3 +4$ $du = 6x^2dx \rightarrow x^2dx=1/6 * du$ $\int\limits_{?}^{?}\sqrt{u}*du/6 = 1/9 * \sqrt{u^3} = 1/9 * \sqrt{(2x^3+4)^3}$ But Wolfram got something else entirely but I fail to see my mistake.

2. hartnn

its 3x^2+4 in sqrt but u put u = 2x^3+4.....

3. TomLikesPhysics

Sry I got the first line wrong. This is the integral I am supposed so solve. $\int\limits_{?}^{?}x^2*\sqrt{2x^3+4} dx$

4. hartnn

i don't see any fault in your work.

5. TomLikesPhysics

Ups sry. There is no mistake. Wolfram just pulls a 2 out of the squareroot term.

6. TomLikesPhysics

Thanks for the douple checking.^^ I am sorry.

7. hartnn

no problem welcome :)