Question

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

Answer 1

Let $I=\int \sqrt{x^2+4x-5}dx$

=$\int\sqrt{(x^2+4x+4)-9}dx$

$=\int \sqrt{(x+2)^2-(3)^2}dx$

It is known that,$\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\mid x+\sqrt{x^2-a^2}\mid+C$

∴$I= \frac{(x+2)}{2}\sqrt{x^2+4x-5}-\frac{9}{2}\log\mid (x+2)+\sqrt{x^2+4x-5}\mid +C$