Question

Integrate the function $\sqrt{x^2+4x-5}$

Answer 1

Let $I=\int \sqrt{x^2+4x-5}dx$
$=\int \sqrt{(x^2+4x+4)-9}dx$
$=\int \sqrt{(x+2{)}^2-(3{)}^2}dx$
We know that, $\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log |x+\sqrt{x^2-a^2}|+C$
$\therefore I=\frac{(x+2)}{2}\sqrt{x^2+4x-5}-\frac{9}{2}\log |(x+2)+\sqrt{x^2+4x-5}|+C$