# Integral: have madness, need method

• March 7th 2009, 12:29 PM
TYTY
$
\frac{x - 3}{x^2 + 1}
$

This looks like a tan^-1x type answer but the actual answer is different and I am not sure of the process to get there.
• March 7th 2009, 12:49 PM
Mentia
okay so separate it out:

$\frac{ x-3}{x^{2}+1 } = \frac{x }{ x^2+1} - \frac{ 3}{x^2+1 }$

then

$
\int_{ }^{ } \frac{ x-3}{x^{2}+1 }dx = \int_{ }^{ } \frac{x }{ x^2+1}dx - \int_{ }^{ } \frac{ 3}{x^2+1 }dx$

The part with the 3 is just arctan. Use a substitution on the other part:
u = x^2 + 1 --> du = 2xdx ---> dx = du/(2x), and so the integral should give a natural log.
• March 7th 2009, 12:52 PM
TYTY
Quote:

Originally Posted by Mentia
Whoops, that funny part is:

$
\int_{ }^{ } \frac{ x-3}{x^{2}+1 }dx = \int_{ }^{ } \frac{x }{ x^2+1}dx - \int_{ }^{ } \frac{ 3}{x^2+1 }dx$

haha thank goodness I didn't have to try to sort out the first post lol. This is so obvious no that I see it :o

thank you so much for taking the time