10x(5x^2 - 3)^4 dx

I got it to [10x (5x^2)^5]/5

First off, is this correct? And if so, should it be simplified?

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- Aug 10th 2010, 07:51 PMbobsanchezEvaluating the integral:
10x(5x^2 - 3)^4 dx

I got it to [10x (5x^2)^5]/5

First off, is this correct? And if so, should it be simplified? - Aug 10th 2010, 07:57 PMeumyang
Not quite.

$\displaystyle \displaystyle \int 10x(5x^2 - 3)^4 \,dx$

$\displaystyle \begin{aligned}

u &= 5x^2 - 3 \\

du &= 10x \,dx

\end{aligned}$

$\displaystyle \begin{aligned}

\displaystyle \int u^4 \,du &= \dfrac{u^5}{5} + C \\

&= \dfrac{1}{5}(5x^2 - 3)^5 + C

\end{aligned}$ - Aug 10th 2010, 08:32 PMbobsanchez
Okay, so if the 10x is already there you don't have to put it back in the problem?

- Aug 10th 2010, 08:57 PMeumyang
That's right, you don't.

- Aug 10th 2010, 08:59 PMbobsanchez
Okay, thanks guys. That probably explains some of my other problems. I really appreciate it.