Question

Evaluate $\int \frac{dx}{x+4-x^2}$

Answer 1

$\int \frac{dx}{x+4-x^2}=-\int \frac{dx}{x^2-x-4}$

$=\int \frac{-dx}{{(x-\frac{1}{2})}^2-4-\frac{1}{4}}$

$=\int \frac{-dx}{{(x-\frac{1}{2})}^2-{\left(\frac{\sqrt{17}}{2}\right)}^2}$

$=\frac{-1}{\sqrt{17}}ln\left(\frac{x-\frac{1}{2}-\frac{\sqrt{17}}{2}}{x-\frac{1}{2}+\frac{\sqrt{17}}{2}}\right)+c$

$=\frac{-1}{\sqrt{17}}ln\left(\frac{2x-1-\sqrt{17}}{2x-1+\sqrt{17}}\right)+c$