How can we solve this integral $\frac{x\,dx}{x^2-x-6}$?

Question

$$\int_0^1\frac{x\, dx}{x^2-x-6} = [uv]^1_0 -\int_0^1 v\,du = ?$$

$u=x,du=dx,dv=\frac{dx}{x^2-x-6},v=?$ (this is my problem.)

Thanks for your help. :)

Answer 1

HINT:

Write $x^2-x-6$ as $(x-3)(x+2)$ and use partial fractions: $$\frac 1{x^2-x-6}=\frac A{x-3}+\frac B{x+2}$$ where $A,B$ are constants. Then multiply this by $x$ so that the integral can be easily solved using $\ln$.

Answer 2

Find constants $A,\,B$ such that $\frac{x}{x^2-x-6}=\frac{A}{x-3}+\frac{B}{x+2}$. I get $A=\frac{3}{5},\,B=\frac{2}{5}$.