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$\int \frac{x^9}{\left(4 x^2+1\right)^6} d x$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{10}\left(\frac{1}{\mathrm{x}^2}+4\right)^{-5}+\mathrm{C}$
$\frac{1}{10}\left(\frac{1}{\mathrm{x}^2}+4\right)^{-5}+\mathrm{C}$
We have, $I=\int \frac{x^9}{\left(4 x^2+1\right)^6} d x$
$$
=\int \frac{x^9 d x}{x^{12}\left(4+\frac{1}{x^2}\right)^6}=\int \frac{d x}{x^3\left(4+\frac{1}{x^2}\right)^6}
$$
Put $4+\frac{1}{x^2}=t \Rightarrow \frac{-2}{x^3} d x=d t$
$$
\begin{aligned}
I &=-\frac{1}{2} \int \frac{d t}{t^6}=\frac{-1}{2} \int t^{-6} d t \\
&=-\frac{1}{2} \frac{t^{-5}}{-5}+C=\frac{1}{10}\left(4+\frac{1}{x^2}\right)^{-5}+C
\end{aligned}
$$
$$
=\int \frac{x^9 d x}{x^{12}\left(4+\frac{1}{x^2}\right)^6}=\int \frac{d x}{x^3\left(4+\frac{1}{x^2}\right)^6}
$$
Put $4+\frac{1}{x^2}=t \Rightarrow \frac{-2}{x^3} d x=d t$
$$
\begin{aligned}
I &=-\frac{1}{2} \int \frac{d t}{t^6}=\frac{-1}{2} \int t^{-6} d t \\
&=-\frac{1}{2} \frac{t^{-5}}{-5}+C=\frac{1}{10}\left(4+\frac{1}{x^2}\right)^{-5}+C
\end{aligned}
$$
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