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

$\sqrt{x^2-4x-5} + 4\ln|x-2 + \sqrt{x^2-4x-5}| + C$

The correct answer is Option (1) → $\sqrt{x^2-4x-5} + 4\ln|x-2 + \sqrt{x^2-4x-5}| + C$

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

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

Let $u = x-2$, then $x = u+2$ and $du = dx$

$I = \int \frac{u+4}{\sqrt{u^2-9}} du$

$I = \int \frac{u}{\sqrt{u^2-9}} du + \int \frac{4}{\sqrt{u^2-9}} du$

$I = I_1 + I_2$

Now, $I_1= \int \frac{u}{\sqrt{u^2-9}} du$

Let $v = u^2-9$, then $dv = 2u du$

$I_1 = \frac{1}{2} \int \frac{1}{\sqrt{v}} dv = \frac{1}{2} \int v^{-\frac{1}{2}} dv$

$=\frac{1}{2}. 2v^{\frac{1}{2}} = \sqrt{v} = \sqrt{u^2-9}$

And $I_2=\int \frac{4}{\sqrt{u^2-9}} du $

$4 = \int \frac{1}{\sqrt{u^2-3^2}} du$

$ = 4 \ln |u + \sqrt{u^2-9}|$

Therefore, $I = \sqrt{u^2-9} + 4 \ln |u + \sqrt{u^2-9}| + C$

Substitute $u = x-2$, we get

$I = \sqrt{(x-2)^2-9} + 4 \ln |x-2 + \sqrt{(x-2)^2-9}| + C$

$I = \sqrt{x^2-4x-5} + 4 \ln |x-2 + \sqrt{x^2-4x-5}| + C$