1. integral with improper fraction

The integral of $\displaystyle x^3/\sqrt(1+x)$

How do you start this integral since it is an improper fraction. I tried squaring the top and bottom, but that didn't seem to work. What else should I try

2. Originally Posted by stevo970
The integral of $\displaystyle x^3/\sqrt(1+x)$

How do you start this integral since it is an improper fraction. I tried squaring the top and bottom, but that didn't seem to work. What else should I try
A substitution. Let $\displaystyle u^2 = 1 + x$

Can you finish?

3. correction $\displaystyle x^3/\sqrt(1+x^2)$

4. Originally Posted by stevo970
correction $\displaystyle x^3/\sqrt(1+x^2)$
That's even better! we won't have to cube a binomial.

Similar approach, let $\displaystyle u^2 = 1 + x^2$

5. so would you then have to figure out

du=$\displaystyle x/\sqrt(1+x^2)$dx or is that not the next step

6. Originally Posted by stevo970
so would you then have to figure out

du=$\displaystyle x/\sqrt(1+x^2)$dx or is that not the next step
That's not the next step...

$\displaystyle u^2 = 1 + x^2 \implies \boxed{x^2 = u^2 - 1}$

$\displaystyle \Rightarrow 2u~du = 2x~dx$

$\displaystyle \Rightarrow u~du = x~dx$

Now substitute and finish up