If $\int \frac{dx}{(x-1)^3/^4. (x+2)^5/^4} = \alpha[1 - g(x)]^\beta + c$, where $c$ is a constant of integration, then which of the following are true?

(A) $\alpha = \frac{2}{3}$

(B) $\beta = \frac{3}{4}$

(C) $3\alpha + 4\beta = 5$

(D) $g(x) = \frac{3}{(x+2)}$

Choose the correct answer from the options given below:

We need to evaluate:

$I = \int \frac{dx}{(x-1)^{3/4}\,(x+2)^{5/4}}$

The total power in the denominator is $\frac{3}{4} + \frac{5}{4} = 2$, so we substitute:

$t = \frac{x-1}{x+2}$

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$dt = \frac{(x+2)(1) - (x-1)(1)}{(x+2)^2}\,dx$

$= \frac{3}{(x+2)^2}\,dx$

$dx = \frac{(x+2)^2}{3}\,dt$

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Since $x - 1 = t(x+2)$, we get $(x-1)^{3/4} = t^{3/4}(x+2)^{3/4}$

The denominator becomes:

$t^{3/4}(x+2)^{3/4} \cdot (x+2)^{5/4} = t^{3/4}(x+2)^{2}$

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$I = \int \frac{(x+2)^2/3}{t^{3/4}\,(x+2)^{2}}\,dt$

$= \frac{1}{3}\int t^{-3/4}\,dt$

$= \frac{1}{3} \cdot \frac{t^{1/4}}{1/4} + c$

$= \frac{4}{3}\,t^{1/4} + c$

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$I = \frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1/4} + c$

$\frac{x-1}{x+2} = \frac{(x+2)-3}{x+2} = 1 - \frac{3}{x+2}$

$I = \frac{4}{3}\left(1 - \frac{3}{x+2}\right)^{1/4} + c$

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Comparing with $\alpha\,[1 - g(x)]^{\beta} + c$:

$\alpha = \frac{4}{3}$, $\quad \beta = \frac{1}{4}$, $\quad g(x) = \frac{3}{x+2}$

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(A) $\alpha = \frac{2}{3}$ → $\alpha = \frac{4}{3} \neq \frac{2}{3}$ ❌

(B) $\beta = \frac{3}{4}$ → $\beta = \frac{1}{4} \neq \frac{3}{4}$ ❌

(C) $3\alpha + 4\beta = 3 \times \frac{4}{3} + 4 \times \frac{1}{4}$

$= 4 + 1 = 5$ ✅

(D) $g(x) = \frac{3}{x+2}$ ✅

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The correct answer is (C) and (D) only.