Evaluate the integral ∫ [x^3/sqrt(x^2+1)] in two different ways.
a) Integration by parts
b) The substitution u = sqrt(x^2+1)
Not sure how to do this, help would be greatly appreciated!
$\displaystyle \begin{aligned}& u: = \sqrt{x^2+1} \Rightarrow \frac{du}{dx} = \frac{2x}{2\sqrt{x^2+1}} = \frac{x}{\sqrt{x^2+1}} \Rightarrow dx = \frac{\sqrt{x^2+1}}{x}\;{du} \\& \therefore \int\frac{x^3}{\sqrt{x^2+1}}\;{dx} = \int\frac{x^3\sqrt{x^2+1}}{x\sqrt{x^2+1}}\;{du} = \int x^2\;{du} = \int (u^2-1)\;{du} = \cdots \end{aligned}$