$\int \tan (x - α) \tan (x + α) \tan 2 x d x =$

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$\frac{1}{2} \log |\sec 2 x| - \log |\sec (x + α)| - \log |\sec (x - α)| + C$

$\log |\sqrt{\sec 2 x}| + \log |\sec (x + α)| + \log |\sec (x + α)| + C$

$\frac{1}{\tan (x - α)} + \frac{1}{\tan (x + α)} + \frac{1}{\tan 2 x} + C$

$\frac{1}{2} \tan (x - α) \tan (x + α) + \tan 2 x + C$

answer is D.

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Detailed Solution

$\begin{array}{l} \underline{I = \int \tan (x - α) \tan (x + α) \tan 2 x d x} \\ Now \tan 2 x = \tan (x + α + x - α) \\ \tan 2 x = \frac{\tan (x + α) + \tan (x - α)}{1 - \tan (x + α) \tan (x - α)} \\ \tan 2 x (1 - \tan (x + α) \tan (x - α) = \tan (x + α) + \tan (x - α) \\ ∴ \tan (x - α) \tan (x + α) \tan 2 x = \tan 2 x - \tan (x + α) - \tan (x - α) \\ \Rightarrow I = \int \tan (x - α) \tan (x + α) \tan 2 x d x \\ = \int {\tan 2 x - \tan (x + α) - \tan (x - α)} d x \\ = \frac{1}{2} \log | \sec 2 x | - \log | \sec (x + α) | - \log | \sec (x - α) | + C \end{array}$