I don't understand this u-sub step..

Question

Solution:

I've been staring at this step for an hour now.. I really can't understand how he changes $x^3$ into $u^{3/2}$ with the selection of $u = 9x^2 - 1$...

Can anybody help me with this? Thanks...

Answer 1

$I=\int x^3 \sqrt {9x^2-1}\,dx $

Let $\, u= {9x^2-1} \implies x^2 = \frac{u+1}{9} \\du = 18x\,dx \implies\,dx = \frac{1}{18x}du$

$I=\int\frac{u+1}{9}\sqrt{9(\frac{u+1}{9})-1}\,.\frac{1}{18}\,du $

$I=\frac{1}{162}\large\int(u+1)\sqrt u\,du$

$I=\,\frac{1}{162}\large\int u^{\frac32} + \sqrt u\,\,du$

Answer 2

$$\int x^3\sqrt{9x^2-1}\, dx$$

$$u = 9x^2-1$$

$$9x^2=u+1$$

\begin{align} \int x^3 \sqrt{9x^2-1}\, dx &= \int x^3\sqrt{u} \frac{1}{18x}\, du \\ &=\frac{1}{18} \int x^2 \sqrt{u}\, du \\ &=\frac{1}{18} \int \frac{u+1}{9} \sqrt{u}\, du \\ \end{align}