# Integrate the functions in Exercises 1 to 24.    Q16.    $e^{3\log x} (x^4 + 1)^{-1}$

H Harsh Kankaria

Given the function to be integrated as

$e^{3\log x} (x^4 + 1)^{-1}$
$= e^{\log x^3}(x^4 + 1)^{-1} = \frac{x^3}{x^4 + 1}$

Let $I = \int e^{3\log x} (x^4 + 1)^{-1}$

Let $x^4 = t \implies 4x^3 dx = dt$

$I = \int e^{3\log x} (x^4 + 1)^{-1} = \int \frac{x^3}{x^4 + 1}$

$\implies I = \frac{1}{4}\log(x^4 +1) + C$

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