Question

I need some help with an Integral (Substitution).

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

i don't see any fault in your work.

Answer 5

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

Answer 6

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

Answer 7

no problem
welcome :)