Question

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

Answer 1

$\int \sqrt{x^2+4x-5}.dx$

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

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

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

Using

$\int \sqrt{(x^2-a^2)}.dx=\frac{1}{2}x\sqrt{(x^2-a^2)}-\frac{a^2}{2}\log |x+\sqrt{(x^2-a^2)}|$

$=\frac{(x+2)}{2}\sqrt{(x+2{)}^2-(3{)}^2}-\frac{(3{)}^2}{2}\log |x+2+\sqrt{(x+2{)}^2-(3{)}^2}|+C$

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

Where $C$ is constant of integration