can someone solve
∫ (4dx)/(x²-2x-3)
i don't think this equation diverges i think the denominator is 10 and not 4 but i am not sure
$\displaystyle \frac{4}{(x+1)(x-3)} = \frac{A}{x+1} + \frac{B}{x-3}$
$\displaystyle 4 = A(x-3) + B(x+1)$
let $\displaystyle x = 3$ ... $\displaystyle B = 1$
let $\displaystyle x = -1$ ... $\displaystyle A = -1$
$\displaystyle \frac{4}{(x+1)(x-3)} = \frac{1}{x-3} -\frac{1}{x+1}$
now you can integrate the right side ...