# integral with improper fraction

• Dec 5th 2009, 05:59 PM
stevo970
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
• Dec 5th 2009, 06:01 PM
Jhevon
Quote:

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?
• Dec 5th 2009, 06:01 PM
stevo970
correction $\displaystyle x^3/\sqrt(1+x^2)$
• Dec 5th 2009, 06:03 PM
Jhevon
Quote:

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 $\displaystyle u^2 = 1 + x^2$
• Dec 6th 2009, 02:07 PM
stevo970
so would you then have to figure out

du=$\displaystyle x/\sqrt(1+x^2)$dx or is that not the next step
• Dec 6th 2009, 02:24 PM
Jhevon
Quote:

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