int {{e^{3log x}}{{({x^4} + 1)}^{ - 1}},dx} = | vedclass

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(B) Given integral: $I = \int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}dx}$Using the property $n\log x = \log(x^n)$ and $e^{\log f(x)} = f(x)$,we have:$e^{

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

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(B) Given integral: $I = \int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}dx}$Using the property $n\log x = \log(x^n)$ and $e^{\log f(x)} = f(x)$,we have:$e^{3\log x} = e^{\log(x^3)} = x^3$So,the integral becomes:$I = \int \frac{x^3}{x^4 + 1} dx$Let $u = x^4 + 1$. Then $du = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} du$.Substituting these into the integral:$I = \int \frac{1}{u} \cdot \frac{1}{4} du = \frac{1}{4} \int \frac{1}{u} du$$I = \frac{1}{4} \log|u| + c$Substituting back $u = x^4 + 1$:$I = \frac{1}{4} \log(x^4 + 1) + c$

$\int {{e^{3\log x}}{{({x^4} + 1)}^{ - 1}}\,dx} = $

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