The value of $\int \frac{\mathrm{d} x}{(x+1)^{3 / 4}(x-2)^{5 / 4}}$ is equal to
The value of $\int \frac{\mathrm{d} x}{(x+1)^{3 / 4}(x-2)^{5 / 4}}$ is equal to
- $4\left(\frac{x+1}{x-2}\right)^{1 / 4}+\mathrm{c}$, where c is a constant of integration.
- $\quad 4\left(\frac{x-2}{x-1}\right)^{1 / 4}+\mathrm{c}$, where c is a constant of integration.
- $\frac{-4}{3}\left(\frac{x-2}{x+1}\right)^{1 / 4}+\mathrm{c}$, where c is a constant of integration.
- $\frac{-4}{3}\left(\frac{x+1}{x-2}\right)^{1 / 4}+\mathrm{c}$, where c is a constant of integration.
Solution
$\begin{aligned} & \text {Let } \mathrm{I}=\int \frac{\mathrm{d} x}{(x+1)^{\frac{3}{4}} \cdot(x-2)^{\frac{5}{4}}} \\ & \quad=\int \frac{\mathrm{d} x}{\left(\frac{x+1}{x-2}\right)^{\frac{3}{4}}(x-2)^2} \\ & \begin{aligned} \text {Put } & \frac{x+1}{x-2}=\mathrm{t} \Rightarrow \frac{-3}{(x-2)^2} \mathrm{~d} x=\mathrm{dt} \\ \quad \mathrm{I} & =-\frac{1}{3} \int \frac{\mathrm{dt}}{\mathrm{t}^{\frac{3}{4}}} \\ & =-\frac{1}{3} \cdot \frac{\mathrm{t}^{\frac{1}{4}}}{\frac{1}{4}}+\mathrm{c} \\ & =-\frac{4}{3}\left(\frac{x+1}{x-2}\right)^{\frac{1}{4}}+\mathrm{c}\end{aligned}\end{aligned}$
Asked in: MHT CET 2024 (09 May Shift 2)
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