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$\int e^{3 \log x}\left(x^4+1\right)^{-1} d x$ is equal |
$\log \left(x^4+1\right)+C$ $\frac{1}{4} \log \left(x^4+1\right)+C$ $-\log \left(x^4+1\right)+C$ none of these |
$\frac{1}{4} \log \left(x^4+1\right)+C$ |
We have, $I =\int e^{3 \log x}\left(x^4+1\right)^{-1} d x=\int \frac{e^{\log x^3}}{x^4+1} d x=\int \frac{x^3}{x^4+1} d x$ $\Rightarrow I =\frac{1}{4} \int \frac{1}{x^4+1} d\left(x^4+1\right)=\frac{1}{4} \log \left(x^4+1\right)+C$ |
