integral with improper fraction

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December 5th 2009, 05:59 PM
stevo970

integral with improper fraction

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

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
December 5th 2009, 06:01 PM
Jhevon

Quote:

Originally Posted by stevo970

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

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 $u^2 = 1 + x$

Can you finish?
December 5th 2009, 06:01 PM
stevo970

correction $x^3/\sqrt(1+x^2)$
December 5th 2009, 06:03 PM
Jhevon

Quote:

Originally Posted by stevo970

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

That's even better! we won't have to cube a binomial.

Similar approach, let $u^2 = 1 + x^2$
December 6th 2009, 02:07 PM
stevo970

so would you then have to figure out

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

Quote:

Originally Posted by stevo970

so would you then have to figure out

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

That's not the next step...

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

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

$\Rightarrow u~du = x~dx$

Now substitute and finish up

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