Why does ∫x(2-x^2)^3 dx not equal ∫du where u=5-x^2?

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Why does ∫x(2-x^2)^3 dx not equal ∫du where u=5-x^2?

Mathematics
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...you mean u=2-x^2? Because after you find du = -2x dx, you have to solve that differential equation for dx to sub back into the integral: dx = du/(-2x). Put that expression instead of dx, put 'u' instead of (2-x^2) and the extra 'x' will cancel out.
Then, you'll end up with -1/2 * ∫ u^3 du. Evaluate, and when you have the finished expression you can plug "u" back into it.
I'm sorry, the 2 in the integral was supposed to be a 5

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But...the same concept applies. Thank you again for all of your help Quantum! :)
First need to expand x(2-x^2)^3 which equals -x^7 + 6x^5 -12 x^3 + 8x.Then you integrate the function. Or you can ... u = (2-x^2) du = -2xdu -du/2 = dx \[- 1/2\int\limits_{?}^{?} u^3 du =(-1/2) u^4/4 =- (2-x^2)^4/8 + C\]
Thank you so much Kynosis...excellent structure. I really appreciate it.

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