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BITSAT 2022

Exam: BITSAT
Year: 2022
Subject: Mathematics
Chapter: Indefinite Integration

BITSAT 2022 · Mathematics · Indefinite Integration

The value of ∫ 1 [(x-1)^3(x+2)^5]^ (1)/(4) d x , is

The value of

\int \frac{1}{\left[(x-1)^3(x+2)^5\right]^{\frac{1}{4}}} d x

, is

Options

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

Correct answer

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

Step-by-step solution

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

Let

\frac{x-1}{x+2}=t \Rightarrow \frac{3 d x}{(x+2)^2}=d t

\begin{aligned} \Rightarrow \quad I & =\frac{1}{3} \int \frac{1}{t^{3 / 4}} d t=\frac{1}{3}\left(\frac{t^{1 / 4}}{1 / 4}\right)+ C \\ & =\frac{4}{3} t^{1 / 4}+C=\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1 / 4}+ C \end{aligned}

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