Integrate the function: $e^{3 \log x}(x^{4}+1)^{-1}$

Given integral is $I = \int e^{3 \log x}(x^{4}+1)^{-1} dx$.
Using the property of logarithms,$n \log x = \log x^{n}$,we have $e^{3 \log x} = e^{\log x^{3}} = x^{3}$.
Thus,the integral becomes $I = \int \frac{x^{3}}{x^{4}+1} dx$.
Let $t = x^{4}+1$. Then $dt = 4x^{3} dx$,which implies $x^{3} dx = \frac{dt}{4}$.
Substituting these into the integral,we get $I = \int \frac{1}{t} \cdot \frac{dt}{4} = \frac{1}{4} \int \frac{1}{t} dt$.
Integrating,we get $I = \frac{1}{4} \log |t| + C$.
Substituting back $t = x^{4}+1$,we get $I = \frac{1}{4} \log |x^{4}+1| + C$.
Since $x^{4}+1 > 0$ for all real $x$,we can write $I = \frac{1}{4} \log (x^{4}+1) + C$.