If \(\displaystyle\int\frac{dx}{(x - 1)^{3/4}\cdot(x + 2)^{5/4}} = \alpha[1 - g(x)]^{\beta} + c\), where c is a constant of integration, then which of the following are true?
(A) \(\alpha =\displaystyle\frac{2}{3}\)
(B) \(\beta =\displaystyle\frac{3}{4}\)
(C) \(3\alpha +4\beta = 5\)
(D) \(g(x)=\displaystyle\frac{3}{(x+2)}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) and (D) only
The powers add up to \(\frac{3}{4} + \frac{5}{4} = 2\), so write the integrand using the ratio \(\frac{x-1}{x+2}\):
\( \displaystyle (x-1)^{3/4}(x+2)^{5/4} = \left(\frac{x-1}{x+2}\right)^{3/4}(x+2)^{2} \)
Let
\( \displaystyle t = \frac{x-1}{x+2} \)
\( \displaystyle dt = \frac{3}{(x+2)^2}\,dx \)
Then
\( \displaystyle \int\frac{dx}{(x-1)^{3/4}(x+2)^{5/4}} = \frac{1}{3}\int t^{-3/4}\,dt \)
\( \displaystyle = \frac{4}{3}\,t^{1/4} + c \)
\( \displaystyle = \frac{4}{3}\left[1 - \frac{3}{x+2}\right]^{1/4} + c \)
(using \(\frac{x-1}{x+2} = 1 - \frac{3}{x+2}\)).
So \(\alpha = \frac{4}{3}\), \(\beta = \frac{1}{4}\) and \(g(x) = \frac{3}{x+2}\).
Checking the statements
(A) \(\alpha = \frac{2}{3}\) — Incorrect. \(\alpha\) is \(\frac{4}{3}\).
(B) \(\beta = \frac{3}{4}\) — Incorrect. \(\beta\) is \(\frac{1}{4}\).
(C) \(3\alpha + 4\beta = 5\) — Correct. \(3 \times \frac{4}{3} + 4 \times \frac{1}{4} = 4 + 1 = 5\).
(D) \(g(x) = \frac{3}{(x+2)}\) — Correct. The bracket is \(1 - \frac{3}{x+2}\), so \(g(x) = \frac{3}{x+2}\).