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TomLikesPhysics
 2 years ago
Best 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.

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

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.0Sry 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
 2 years ago
Best ResponseYou've already chosen the best response.1i don't see any fault in your work.

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

TomLikesPhysics
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks for the douple checking.^^ I am sorry.
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