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TomLikesPhysicsBest ResponseYou've already chosen the best response.0
\[\int\limits_{?}^{?}x^2*\sqrt{3x^2+4}dx\] 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.
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
its 3x^2+4 in sqrt but u put u = 2x^3+4.....
 one year ago

TomLikesPhysicsBest ResponseYou've already chosen the best response.0
Sry I got the first line wrong. This is the integral I am supposed so solve. \[\int\limits_{?}^{?}x^2*\sqrt{2x^3+4} dx\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
i don't see any fault in your work.
 one year ago

TomLikesPhysicsBest ResponseYou've already chosen the best response.0
Ups sry. There is no mistake. Wolfram just pulls a 2 out of the squareroot term.
 one year ago

TomLikesPhysicsBest ResponseYou've already chosen the best response.0
Thanks for the douple checking.^^ I am sorry.
 one year ago
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