integral with improper fraction

**stevo970** · December 5th 2009, 05:59 PM

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

**Jhevon** · December 5th 2009, 06:01 PM

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?

**stevo970** · December 5th 2009, 06:01 PM

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

**Jhevon** · December 5th 2009, 06:03 PM

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 $u^2 = 1 + x^2$

**stevo970** · December 6th 2009, 02:07 PM

so would you then have to figure out

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

**Jhevon** · December 6th 2009, 02:24 PM

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

Thread: integral with improper fraction

LinkBack

Thread Tools

Display

integral with improper fraction

Similar Math Help Forum Discussions

please help improper fraction into partial fraction

partial fraction from improper fraction

Displaying time in Improper fraction form

Improper integral 3

Improper integral 2

Search Tags