Q113 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

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

  • Correct answer
    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
    $\frac{4}{3}\left(\frac{x+2}{x-1}\right)^{\frac{1}{4}}+c$
  • D
    $\frac{3}{4}\left(\frac{x-2}{x-1}\right)^{\frac{1}{4}}+c$

Solution

Let $I=\int \frac{d x}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}$ $$ \begin{aligned} & =\int \frac{(x-1)^{\frac{-3}{4}} d x}{(x+2)^{\frac{-3}{4}}\left(x+2^2\right)} \\ & =\int\left(\frac{x-1}{x+2}\right)^{\frac{-3}{4}}\left(\frac{1}{x+2}\right)^2 d x \end{aligned} $$ Let $\left(\frac{x-1}{x+2}\right)=t$ $\begin{aligned} & \frac{(x+2)-(x-1)}{(x+2)^2} d x=d t \Rightarrow \frac{d x}{(x+2)^2}=\frac{d t}{3} \\ & \therefore I=\frac{1}{3} \int t^{\frac{-3}{4}} d t \\ & =\frac{4}{3} t^{\frac{1}{4}}+\mathrm{C}=\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+\mathrm{C}\end{aligned}$

Hence, the correct answer is option: 'A'.

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