Question

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

Answer 1

$e^{3\log x}(x^4+1{)}^{-1}=e^{\log x^3}(x^4+1{)}^{-1}=\frac{x^3}{(x^4+1)}$
Let $x^4+1=t\Rightarrow 4x^{3}dx=dt$
$\Rightarrow \int e^{3\log x}(x^4+1{)}^{-1}dx=\int \frac{x^3}{(x^4+1)}dx$
$=\frac{1}{4}\int \frac{dt}{t}=\frac{1}{4}\log |t|+C$
$=\frac{1}{4}\log |x^4+1|+C=\frac{1}{4}\log (x^4+1)+C$