# Thread: Simplifying an Integral

1. ## Simplifying an Integral

How does this
$\displaystyle \int\frac{x^5}{\sqrt[3]{x^2+4}}$

become
$\displaystyle \frac{1}{2}\int\frac{x^4}{\sqrt[3]{x^2+4}}$

According to my notes I let

u = x^2 + 4
x^2 = u - 4

but I do not see how that gives me the above.

2. Originally Posted by gammaman
How does this
$\displaystyle \int\frac{x^5}{\sqrt[3]{x^2+4}}$

become
$\displaystyle \frac{1}{2}\int\frac{x^4}{\sqrt[3]{x^2+4}}$

According to my notes I let

u = x^2 + 4
x^2 = u - 4

but I do not see how that gives me the above.
$\displaystyle \int\frac{x^5}{\sqrt[3]{x^2+4}} \, dx$

$\displaystyle u = x^2 + 4$

$\displaystyle du = 2x \, dx$

$\displaystyle x^2 = u - 4$

$\displaystyle \frac{1}{2}\int\frac{x^4}{\sqrt[3]{x^2+4}} \cdot 2x \, dx$

substitute ...

$\displaystyle \frac{1}{2}\int\frac{(u-4)^2}{\sqrt[3]{u}} \, du$

3. put $\displaystyle u^3=x^2+4$ an get: $\displaystyle \frac{3} {2}\int {\frac{{\left( {u^3 - 4} \right)^2 }} {u} \cdot u^2 du}$

4. what happens to the X^5 though? Do we just get rid of it? Why?

5. Originally Posted by gammaman
what happens to the X^5 though? Do we just get rid of it? Why?
it's still there ... what is $\displaystyle \frac{1}{2} \cdot x^4 \cdot 2x$ ?