Question:
$e^{3 \log x}\left(x^{4}+1\right)^{-1}$
Solution:
$e^{3 \log x}\left(x^{4}+1\right)^{-1}=e^{\log x^{3}}\left(x^{4}+1\right)^{-1}=\frac{x^{3}}{\left(x^{4}+1\right)}$
Let $x^{4}+1=t \Rightarrow 4 x^{3} d x=d t$
$\Rightarrow \int e^{3 \log x}\left(x^{4}+1\right)^{-1} d x=\int \frac{x^{3}}{\left(x^{4}+1\right)} d x$
$=\frac{1}{4} \int \frac{d t}{t}$
$=\frac{1}{4} \log |t|+\mathrm{C}$
$=\frac{1}{4} \log \left|x^{4}+1\right|+\mathrm{C}$
$=\frac{1}{4} \log \left(x^{4}+1\right)+\mathrm{C}$