The integral $\int \frac{1}{\sqrt[4]{(x-1)^{3}(x+2)^{5}}} \mathrm{dx}$ is equal to: (where C is a constant of integration)

(C) $\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$

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

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

(D) $\frac{3}{4}\left(\frac{x+2}{x-1}\right)^{\frac{1}{4}}+C$

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Indefinite Integrals

The integral

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

is equal to: (where C is a constant of integration)

Answer: (C)

Write the integrand as

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

. With

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

d t = \frac{3}{( x + 2 ) ^{2}} d x

, so the integral is

\frac{1}{3} \int t^{- 3/4} d t = \frac{1}{3} \cdot \frac{t ^{1/4}}{1/4} = \frac{4}{3} t^{1/4}

. Hence it equals

\frac{4}{3} (\frac{x - 1}{x + 2})^{1/4} + C

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Single correct option