Integral of $\int \frac {ax+b}{x^2-ax-b} dx$

Question

How do I compute the integral $\int \frac {ax+b}{x^2-ax-b} dx$? In the beginning it seemed like it would be something like the logarithm of the denominator, but it's not that simple in the end. Would a trigonometric substitution help in this case?

Answer 1

Hint: The derivative of the denominator is $2x-a$. So , rewrite the numerator as, $$ax+b=\dfrac{a}{2}(2x-a)+b+\dfrac{a^2}{2}$$ Therefore, $$\int \frac{\dfrac{a}{2}(2x-a)+\left(b+\dfrac{a^2}{2}\right)}{x^2-ax-b}\;\mathrm{d}x=\frac{a}{2}\int \frac{2x-a}{x^2-ax-b}\;\mathrm{d}x+\left(b+\dfrac{a^2}{2}\right)\int \frac{1}{x^2-ax-b}\;\mathrm{d}x$$ For the first integral, substitute $x^2-ax-b=t$ and you will see the purpose of the hint and the second integral is standard.