# Integration of Rational Functions By Partial Fractions

• Oct 2nd 2008, 05:57 PM
HellaciousD
Integration of Rational Functions By Partial Fractions
Hello all,

I am trying to integrate (3x^2+7x+3)/(x^2+1)^2
I am down to the integral of (3/(x^2+1))+ the integral of (7x/(x^2+1)^2)+ the integral of (3/(x^2+1)^2) and I don't really know what to do
I believe the next step starts with arctan(3x) but then I don't know what to do with the other stuff.

Thanks,
Jay
• Oct 2nd 2008, 06:13 PM
Nacho
$\displaystyle \int {\frac{{3x^2 + 7x + 3}} {{\left( {x^2 + 1} \right)^2 }}dx} = \int {\frac{{3x^2 + 3}} {{\left( {x^2 + 1} \right)^2 }}dx + \int {\frac{{7x}} {{\left( {x^2 + 1} \right)^2 }}dx} } =$

$\displaystyle 3\int {\frac{1} {{x^2 + 1}}dx} + 7\int {\frac{x} {{\left( {x^2 + 1} \right)^2 }}dx}$

$\displaystyle \begin{gathered} u = x^2 + 1 \Rightarrow du = 2xdx \hfill \\ \hfill \\ = 3\arctan x + \frac{7} {2}\int {\frac{1} {{u^2 }}du = } 3\arctan x + \frac{7} {2} \cdot \frac{{u^{ - 2 + 1} }} {{ - 2 + 1}} \hfill \\ \end{gathered}$

$\displaystyle = \boxed{3\arctan x + \frac{7} {2} \cdot \frac{1} {{x^2 + 1}}}$

I forget the integration constant :P
• Oct 2nd 2008, 06:21 PM
HellaciousD
Thanks, but the program I'm using isn't accepting that answer. Are you sure it's right?
Jay
• Oct 2nd 2008, 06:25 PM
Nacho
Quote:

Originally Posted by HellaciousD
Thanks, but the program I'm using isn't accepting that answer.
Jay

look that I make a mistake in a sign, before 7/2... should to have a "-"

sorry
• Oct 2nd 2008, 07:20 PM
HellaciousD
Can someone tell me why the integral of 32dx/(x(x^2+4)^2) doesn't come out to 2ln(abs(x))?

Thanks,
Jay