Value of $int\frac{2}{(x-3)\sqrt{x+1}} dx$ is: (Here C is an arbitrary constant)

$\log\left|\frac{\sqrt{x-1}-2}{\sqrt{x+1}+2}\right|+C$

The correct answer is Option (4) → $\log\left|\frac{\sqrt{x-1}-2}{\sqrt{x+1}+2}\right|+C$

Evaluate:

$\displaystyle \int \frac{2}{(x - 3) \sqrt{x + 1}} \, dx$

Substitute:

$t = \sqrt{x + 1} \Rightarrow x = t^2 - 1$, so $dx = 2t \, dt$

Rewrite integral in terms of $t$:

$\int \frac{2}{(t^2 - 1 - 3) \cdot t} \cdot 2t \, dt = \int \frac{4t}{(t^2 - 4) t} \, dt = \int \frac{4}{t^2 - 4} \, dt$

Simplify denominator:

$t^2 - 4 = (t - 2)(t + 2)$

Use partial fractions:

$\frac{4}{(t - 2)(t + 2)} = \frac{A}{t - 2} + \frac{B}{t + 2}$

Multiply both sides by $(t - 2)(t + 2)$:

$4 = A (t + 2) + B (t - 2)$

Set $t = 2$:

$4 = A (4) + B (0) \Rightarrow A = 1$

Set $t = -2$:

$4 = A (0) + B (-4) \Rightarrow B = -1$

Therefore:

$\int \frac{4}{t^2 - 4} dt = \int \left( \frac{1}{t - 2} - \frac{1}{t + 2} \right) dt = \ln|t - 2| - \ln|t + 2| + C = \ln \left| \frac{t - 2}{t + 2} \right| + C$

Substitute back $t = \sqrt{x + 1}$:

${ \int \frac{2}{(x - 3) \sqrt{x + 1}} \, dx = \ln \left| \frac{\sqrt{x + 1} - 2}{\sqrt{x + 1} + 2} \right| + C }$