Let $\text{I}=\int\frac{\tan^2\text{x}\sec^2\text{x}}{1-\tan^6\text{x}}\text{dx}$
Let $\tan^3\text{x}=\text{t}$
$3\tan^2\text{x}\sec^2\text{x}\ \text{dx}=\text{dt}$
$\Rightarrow\text{I}=\int\frac{1}{3}\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{3}\int\frac{\text{dt}}{1-\text{t}^2}$
$=\frac{1}{6}\text{ln}\Big|\frac{1+\text{t}}{1-\text{t}}\Big|+\text{c},$
$\text{I}=\frac{1}{6}\text{ln}\Big|\frac{1+\tan^3\text{x}}{1-\tan^3\text{x}}\Big|+\text{c}$
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