Evaluate this integral:

Question

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

ganeshie8 · Answer

try $u = 3+x^2$ instead

anonymous · Answer

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?

ganeshie8 · Answer

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

ganeshie8 · Answer

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

ganeshie8 · Answer

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

anonymous · Answer

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

ganeshie8 · Answer

yw :)