1. ## Integration Question

In this question I do not see how x^5 becomes an x^4 in the second step.
I understand that x^2 = u-4 and that x^2 * x^2 = x^4, but if we started with x^5 and we now have x^4, were is the extra x?

x^5/ third root of x^2+4

2. Originally Posted by gammaman
In this question I do not see how x^5 becomes an x^4 in the second step.
I understand that x^2 = u-4 and that x^2 * x^2 = x^4, but if we started with x^5 and we now have x^4, were is the extra x?

x^5/ third root of x^2+4
the "extra" x is part of "du" ...

$\int \frac{x^5}{\sqrt[3]{x^2+4}} \, dx$

$u = x^2 + 4$

$du = 2x \, dx$

$x^2 = u - 4$

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

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

$\frac{1}{2} \int \frac{u^2 - 8x + 16}{\sqrt[3]{u}} \, du$

$\frac{1}{2} \int u^\frac{5}{3} - 8x^{\frac{2}{3}} + 16x^{-\frac{1}{3}} \, du$

integrate and back substitute ...