If ∫dx/((x − 1)^3/4.(x + 2)^5/4) = α[1 − g(x)]^β + c, where c is a co…

The correct answer is Option (4) → (C) and (D) only

$\int \frac{dx}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}} =\alpha[1-g(x)]^{\beta}+c$

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

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

$\Rightarrow$ (D) is true

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

$[1-g(x)]^{\beta}=\frac{(x-1)^{\beta}}{(x+2)^{\beta}}$

Differentiating

$\frac{d}{dx}\left[(1-g(x))^{\beta}\right] =\beta(1-g(x))^{\beta-1}\cdot g'(x)(-1)$

$g'(x)=\frac{-3}{(x+2)^2}$

$\Rightarrow \frac{d}{dx}\left[(1-g(x))^{\beta}\right] =\beta\frac{(x-1)^{\beta-1}}{(x+2)^{\beta-1}}\cdot\frac{3}{(x+2)^2}$

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

Compare with integrand

$\frac{1}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}$

So

$\beta-1=-\frac{3}{4}$ and $\beta+1=\frac{5}{4}$

$\beta=\frac{1}{4}$ from both

Hence (B) is false

Now coefficient:

$\alpha\cdot 3\beta =1$

$\alpha\cdot 3\cdot\frac{1}{4}=1$

$\alpha=\frac{4}{3}$

Hence (A) is false

$3\alpha+4\beta=3\cdot\frac{4}{3}+4\cdot\frac{1}{4}=4+1=5$

Hence (C) is true

Correct options: (C) and (D)