Computing the integral $\int \frac{u}{b - au - u^2}\mathrm{d}u$ After working on an ODE I find I am needing to solve the integral $$\int \frac{u}{b - au - u^2}\mathrm{d}u$$ Trig subs, banging heads against walls, and sobbing have not yielded a solution. Yet. Could use a hand, thanks.

Another approach is to use partial fraction

> $$ \frac{u}{b-au-u^2} = \frac{A}{u-\alpha}+\frac{B}{u-\beta} $$

where $\alpha, \beta$ are the roots of $ b-au-u^2 $. You need to determine $A$ and $B$. The answer will have the form

> $$ I = A\ln(u-\alpha)+B\ln(u-\beta)+C. $$

**Note:** Here are the roots

> $$ \alpha = -\frac{a}{2}+\frac{\sqrt {{a}^{2}+4\,b}}{2},\quad \beta = -\frac{a}{2}-\frac{\sqrt {{a}^{2}+4\,b}}{2} $$