I'm stumped on these:
The integral of x^3 (1+x^2)^3/2
The integral of the square root of (x^2 - 1)dx/x^3
The integral of x^3 dx/(4-x^2)^2
do a substitution, $\displaystyle u = 1 + x^2$
note that $\displaystyle \frac {x^2 - 1}{x^3} = \frac {x^2}{x^3} - \frac 1{x^3} = \frac 1x - x^{-3}$The integral of the square root of (x^2 - 1)dx/x^3
another substitution problem, let $\displaystyle u = 4 - x^2$The integral of x^3 dx/(4-x^2)^2
did you try it?
$\displaystyle \int {x^3}{(1 + x^2)^{3/2}}~dx$
Let $\displaystyle u = 1 + x^2 \implies \boxed{x^2 = u - 1}$ ....note that $\displaystyle x^3 = x \cdot x^2$
$\displaystyle \Rightarrow du = 2x~dx$
$\displaystyle \Rightarrow \frac 12 ~du = x~dx$
Thus our integral becomes
$\displaystyle \frac 12 \int \frac {u - 1}{u^{3/2}}~du$
i suppose you can take it from here. the other integral is similar
Well let me edit that--
Trig subs are NOT required, what I meant to say is that I came to my math teacher with some of the work that Jhevon got me started on (thanks again), and he instructed me to also try to work them out using a trig substitution, so i'm back here. Sorry for the confusion