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Homework Statement
I was intent on finding a general formula for every case where the constants are real in this integral:
[tex]\int\frac{dx}{ax^2+bx+c}[/tex]
But thought I would make things progressive by tackling a seemingly easier problem, mainly:
[tex]\int\frac{ax+b}{cx^2+dx+e}dx[/tex]
The Attempt at a Solution
I let [tex]cx^2+dx+e=u[/tex]
[tex]du=(2cx+d)dx[/tex]
Now I needed the numerator to be equivalent to du, so after some manipulation I get
[tex]ax+b=\frac{a}{2c}(2cx+d+\frac{2bc}{a}-d)[/tex]
Which made me realize that this problem isn't any easier than the first, since I'm going to have to solve the first anyway...
[tex]\int\frac{ax+b}{cx^2+dx+e}dx[/tex]
[tex]=\frac{a}{2c}\left(\int\frac{2cx+d}{u}dx+\int\frac{\frac{2bc}{a}-d}{u}dx\right)[/tex]
[tex]=\frac{a}{2c}\left(\int\frac{du}{u}+\frac{2bc}{a}\int\frac{dx}{cx^2+dx+e}-d\int\frac{dx}{cx^2+dx+e}\right)[/tex]
So I guess my question is how do I deal with this integral in the case that the quadratic has no real factors. i.e. b2-4ac<0.
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