Explanation

(b)
Let $I = \int_{}^{}{e^{3\log x}\left( x^{4} + 1 \right)^{- 1}\ dx}$
$= \int_{}^{}{e^{\log x^{3}}\left( x^{4} + 1 \right)^{- 1}dx}$
$= \int_{}^{}{x^{3}\left( x^{4} + 1 \right)^{- 1}dx = \frac{1}{4}\int_{}^{}{\frac{4x^{3}}{(x^{4} + 1)}dx}}$
Let $x^{4} + 1 = t\ \Rightarrow 4x^{3}dx = dt$
$\therefore I = \frac{1}{4}\int_{}^{}{\frac{1}{t}dt = \frac{1}{4}\log{t + c}}$
$\Rightarrow I = \frac{1}{4}\log{\left( x^{4} + 1 \right) + c}$