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I need some help with an Integral (Substitution).

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

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i don't see any fault in your work.
Ups sry. There is no mistake. Wolfram just pulls a 2 out of the squareroot term.
Thanks for the douple checking.^^ I am sorry.
no problem welcome :)

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