I need some help with an Integral (Substitution).

Question

anonymous · Answer

$$\int\limits_{?}^{?}x^2*\sqrt{3x^2+4}dx$$
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.

hartnn · Answer

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

anonymous · Answer

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

hartnn · Answer

i don't see any fault in your work.

anonymous · Answer

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

anonymous · Answer

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

hartnn · Answer

no problem
welcome :)