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Integrate the function
$\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=I(s a y)$
$\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=I(s a y)$
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Verified Answer
$\mathrm{I}=\int \frac{\mathrm{x}^3}{\mathrm{x}^4+1} \mathrm{dx}$, Put $\mathrm{x}^4+1=\mathrm{t} \Rightarrow 4 \mathrm{x}^3 \mathrm{dx}=\mathrm{dt}$
$\therefore \mathrm{I}=\frac{1}{4} \int \frac{\mathrm{dt}}{\mathrm{t}}=\frac{1}{4} \log |\mathrm{t}|+\mathrm{c}=\frac{1}{4} \log \left|\mathrm{x}^4+1\right|+\mathrm{c}$
$\therefore \mathrm{I}=\frac{1}{4} \int \frac{\mathrm{dt}}{\mathrm{t}}=\frac{1}{4} \log |\mathrm{t}|+\mathrm{c}=\frac{1}{4} \log \left|\mathrm{x}^4+1\right|+\mathrm{c}$
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