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The value of $\int \frac{1}{\left[(x-1)^3(x+2)^5\right]^{\frac{1}{4}}} d x$, is
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Verified Answer
The correct answer is:
$\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$
$\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{\frac{1}{4}}+C$
$I=\int \frac{1}{\left(\frac{x-1}{x+2}\right)^{3 / 4}(x+2)^2} d x$
Let $\frac{x-1}{x+2}=t \Rightarrow \frac{3 d x}{(x+2)^2}=d t$
$$
\begin{aligned}
\Rightarrow \quad I & =\frac{1}{3} \int \frac{1}{t^{3 / 4}} d t=\frac{1}{3}\left(\frac{t^{1 / 4}}{1 / 4}\right)+C \\
& =\frac{4}{3} t^{1 / 4}+C=\frac{4}{3}\left(\frac{x-1}{x+2}\right)^{1 / 4}+C
\end{aligned}
$$
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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