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

The integral contains $\sqrt{x+1}$ in the denominator. To simplify, let:

$\sqrt{x+1} = t$

$x + 1 = t^2$

$x = t^2 - 1$

$dx = 2t \, dt$

The term $(x-3)$ becomes:

$x - 3 = t^2 - 1 - 3 = t^2 - 4$

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Substituting into the integral:

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

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

$= \int \frac{4t}{t(t^2-4)} dt$

$= \int \frac{4}{t^2-4} dt$

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The denominator factors as:

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

Using partial fractions:

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

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

For $t = 2$:

$4 = A(4)$

$A = 1$

For $t = -2$:

$4 = B(-4)$

$B = -1$

Therefore:

$\frac{4}{(t-2)(t+2)} = \frac{1}{t-2} - \frac{1}{t+2}$

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Integrating:

$\int \left(\frac{1}{t-2} - \frac{1}{t+2}\right) dt$

$= \ln|t-2| - \ln|t+2| + C$

$= \ln\left|\frac{t-2}{t+2}\right| + C$

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

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