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Question

# Evaluate $\int {e}^{3}\mathrm{log}x{\left({x}^{4}+1\right)}^{-1}dx$

A

${e}^{3\mathrm{log}x}+c$

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B

$\left(\frac{1}{4}\right)\mathrm{log}\left|{x}^{4}+1\right|+c$

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C

$\mathrm{log}\left({x}^{4}+1\right)+c$

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D

$\frac{1}{2}\mathrm{log}\left({x}^{4}+1\right)+c$

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E

$\left(\frac{{x}^{4}}{{x}^{4}+1}\right)+c$

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Solution

## The correct option is B $\left(\frac{1}{4}\right)\mathrm{log}\left|{x}^{4}+1\right|+c$Explanation for the correct option.Finding the integral.Given, $\int {e}^{3}\mathrm{log}x{\left({x}^{4}+1\right)}^{-1}dx$Let, ${x}^{4}+1=t⇒4{x}^{3}dx=dt$Now,${\mathrm{e}}^{3\mathrm{log}\mathrm{x}}{\left({\mathrm{x}}^{4}+1\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\mathrm{e}}^{{\mathrm{logx}}^{3}}{\left({\mathrm{x}}^{4}+1\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{x}}^{3}}{\left({\mathrm{x}}^{4}+1\right)}\left[\because {\mathrm{e}}^{\mathrm{log}\mathrm{x}}=\mathrm{x}\right]$Substituting $t$ in the integral we get,$⇒\int {\mathrm{e}}^{3\mathrm{log}\mathrm{x}}{\left({\mathrm{x}}^{4}+1\right)}^{-1}d\mathrm{x}=\int \frac{{\mathrm{x}}^{3}}{\left({\mathrm{x}}^{4}+1\right)}d\mathrm{x}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\int \mathrm{t}d\mathrm{t}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\mathrm{log}\left|\mathrm{t}\right|+\mathrm{C}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\mathrm{log}\mid {\mathrm{x}}^{4}+1\mid +\mathrm{C}\left[\mathrm{t}={\mathrm{x}}^{4}+1\right]$Hence, option (B) is correct.

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