Correct option is (4) x – α and log_{e}|sin(x – α)|

Detailed Explaination:

\(∫ \frac{tan x + tan α}{tan x - tan α} dx\)

=\(∫ \frac{\frac{\sin x }{\cos x}+ \frac{\sin α}{\cos α}}{\frac{\sin x }{\cos x} - \frac{\sin α}{\cos α}} dx\)

=\(∫ \frac{\frac{\sin x \cos α+ \sin α \cos x}{\cos α .\cos x}}{\frac{\sin x \cos α- \sin α \cos x}{\cos α .\cos x}} dx\)

=\(\int \frac{\sin x \cos α+ \sin α \cos x}{\sin x \cos α-\sin α \cos x} dx\)

\(=\int \frac{\sin( x + α )}{\sin( x - α )} dx\)

\(=\int \frac{\sin( x -α+2α )}{\sin( x - α )} dx\)

\(=\int \frac{\sin( x -α)\cos2α }{\sin( x - α )} dx+\frac{\cos( x -α)\sin2α }{\sin( x - α )} dx\)

\(= (x-\alpha )\cos2\alpha+\sin2\alpha \log_e|sin(x-\alpha)| +C\)