Difficult Integral of a Rational Function

**dgonnella89** · Jun27-09, 03:44 PM

Hey guys I'm wondering if someone could hep me solve this integral. I've been working at it for a few days now (as part of a project I'm doing over the summer) and have gotten stuck. I think I need to make some substitution but I cant see what it is to make.

$LaTeX Code: -\\int\\frac{dI}{I(R+BI+CI^2)}$

I decomposed using partial fractions and reduced it to this:

$LaTeX Code: -\\frac{1}{R}\\int{\\frac{dI}{I}+\\frac{(CI+B)dI}{R+BI+ CI^2}}$

I think I need to make another substitution here for the right-hand part of the integral. Simple U substitution doesn't work but I'm not sure of another method that would help. I tried completing the square for the polynomial on the bottom but that didn't seem to help.

Any help would be really appreciated! Thanks

~~Civilized~~ · Jun27-09, 04:03 PM

**dgonnella89** · Jun27-09, 06:30 PM

Yes I got that answer with mathematica but I need to be able to solve it by hand. Is it possible?

**George Jones** · Jun27-09, 07:14 PM

$LaTeX Code: <BR>\\frac{CI + B}{R + BI + CI^2} = \\frac{CI + B/2}{R + BI + CI^2} + \\frac{B/2}{R + BI + CI^2} = \\frac{1}{2} \\frac{2CI + B}{R + BI + CI^2} + \\frac{B/2}{R + BI + CI^2}<BR>$

Integrate the first term above by inspection or by a simple substitution to get a ln. complete the square for the denominator of the second term to get an arctan.

**dgonnella89** · Jun28-09, 09:44 AM

Ok I was able to do it now thanks!