Difficult Integral of a Rational Function

[dgonnella89]

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.

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

I decomposed using partial fractions and reduced it to this:

-\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]

-\int\frac{d x}{x(R+B x+C x^2)} = -\frac{\frac{2 B \tan ^{-1}\left(\frac{B+2 C x}{\sqrt{4 C
R-B^2}}\right)}{\sqrt{4 C R-B^2}}+\log (x (B+C x)+R)-2 \log (x)}{2
R}

[dgonnella89]

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

[George Jones]

\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}


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]

Ok I was able to do it now thanks!