Question

Evaluate the integral: $\int_{0}^{2}\frac{dx}{x+4-x^2}$

Answer 1

$\int_{0}^{2}\frac{dx}{x+4-x^2}$=$\int_{0}^{2}\frac{dx}{-(-x^2-x-4)}$

=$\int_{0}^{2}\frac{dx}{x^2-x+\frac{1}{4}-\frac{1}{4}-4}$

=$\int_{0}^{2}\frac{dx}{-[(x-\frac{1}{2})^2-\frac{17}{4}]}$

=$\int_{0}^{2}\frac{dx}{(\sqrt{\frac{17}{2}})^2-(\frac{x-1}{2})^2}$

Let x-$\frac{1}{2}$=t $\therefore$ dx=dt

When x=0,t=$-\frac{1}{2}$ and when x=2,t=$\frac{3}{2}$