Question
Integrate the function in Exercise:

$\text{e}^{3}\log\text{x}\big(\text{x}^{4}+1\big)^{-1}$

Answer

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

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