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

.: Let $I = \int {{e^{3\log x}}{{\left( {{x^4} + 1} \right)}^{ - 1}}dx} = \int {{e^{\log {x^3}}}{{\left( {{x^4} + 1} \right)}^{ - 1}}dx}$

$= \int {{x^3}{{\left( {{x^4} + 1} \right)}^{ - 1}}dx} = \int {\cfrac{{{x^3}}}{{{x^4} + 1}}dx}$

Let ${x^4} = t$ $\Rightarrow$ $4{x^3}dx = dt$

$\therefore$ $I = \cfrac{1}{4}\int {\cfrac{{dt}}{{t + 1}}} = \cfrac{1}{4}\log \left( {t + 1} \right) + C = \cfrac{1}{4}log\left( {{x^4} + 1} \right) + C$