int e^{3 log x}left(x^4+1right)^{-1} d x= | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the integral $I = \int e^{3 \log x}(x^4+1)^{-1} dx$. Using the property $e^{\log a} = a$,we have $e^{3 \log x} = e^{\log x^3} = x^3$. So,$I = \int \frac{x^3}{x^4+1} dx$. Let $t = x^4+1$. Then $dt = 4x^3 dx$,which implies $x^3 dx = \frac{1}{4} dt$. Substituting these into the integral,we get $I = \int \frac{1}{4t} dt = \frac{1}{4} \log|t| + C$. Substituting back $t = x^4+1$,we obtain $I = \frac{1}{4} \log(x^4+1) + C$.

$\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=$

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