Question
Evaluate the following integrals:
$\int\frac{3\text{x}^5}{1+\text{x}^{12}}\text{dx}$

Answer

Let $\text{I}=\int\frac{3\text{x}^5}{1+\text{x}^{12}}\text{dx}$
$=\int\frac{3\text{x}^5}{1+(\text{x}^6)^2}\text{dx}$
Let $\text{x}^6=\text{t}$
$\Rightarrow6\text{x}^5\text{dx = dt}$
$\Rightarrow\text{x}^5\text{dx}=\frac{\text{dt}}{6}$
$\text{I}=\frac{3}{6}\int\frac{\text{dt}}{1+\text{t}^2}$
$=\frac{1}{2}\tan^{-1}(\text{t})+\text{C}$ $\Big[\text{Since}\int\frac{1}{\text{x}^2+1}\text{dx}=\tan^{-1}\text{x+C}\Big]$
$\text{I}=\frac{1}{2}\tan^{-1}(\text{x}^6)+\text{C}$

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