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## anonymous one year ago Evaluate this integral:

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1. anonymous

$\int\limits_{}^{} x^{3} * (3+x^{2})^{5/2}$

2. anonymous

So first I set u=x^3 du=3x^2 * dx du * 1/3 = x^2 * dx

3. anonymous

I tried substituting this into the original problem but it didn't work...

4. ganeshie8

try $$u = 3+x^2$$ instead

5. anonymous

Okay so you would get 3+x^2 = u 2xdx = du xdx = du/2 Can you raise both sides of the equation by the third power? Wouldn't it be (xdx)^3 though? dx^3 is not possible?

6. ganeshie8

 3+x^2 = u 2xdx = du xdx = du/2  so far so good

7. ganeshie8

next notice that you can write x^3 as x*x^2

8. ganeshie8

$$3 + x^2 = u\implies x^2 = u-3$$ so the integral becomes \begin{align}\int\limits_{}^{} x^{3} * (3+x^{2})^{5/2}\,dx &= \int\limits_{}^{} x*x^{2} * (3+x^{2})^{5/2}\,dx \\~\\ &=\int (u-3)(u)^{5/2} \frac{du}{2}\end{align}

9. anonymous

ooohoohhhh! I got the answer now! I just didn't see how we could separate x^3 into x*x^2hahaha #simplealgebra... Thank you~

10. ganeshie8

yw :)

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