# Indefinite integrals

• Sep 23rd 2009, 03:40 PM
Yami Sorceress
Indefinite integrals
I'm very confused about this section. Apparently I need to use the substitution rule but I always seem to mess up, especially with fractions. Can anyone help?

Here's an example:

Integral of: x/sqrt x+1

From what I understand you set x+1 = u
you then take the derivative of u which is 1 and set that equal to u'
From here I tend to mess up...
• Sep 23rd 2009, 03:44 PM
mr fantastic
Quote:

Originally Posted by Yami Sorceress
I'm very confused about this section. Apparently I need to use the substitution rule but I always seem to mess up, especially with fractions. Can anyone help?

Here's an example:

Integral of: x/sqrt x+1

From what I understand you set x+1 = u
you then take the derivative of u which is 1 and set that equal to u'
From here I tend to mess up...

$u = x + 1 \Rightarrow \frac{du}{dx} = 1 \Rightarrow dx = du$.

So the integral becomes $\int \frac{u - 1}{\sqrt{u}} \, du = \int \sqrt{u} - \frac{1}{\sqrt{u}} \, du$ and you should be able to integrate each term. Then substitute back $u = x + 1$.