$\int_0^{\pi /2} {\frac{{\tan x}}{{1 + {m^2}{{\tan }^2}x}}} dx$

Let $I = \int_0^{\pi /2} {\frac{{\tan xdx}}{{1 + {m^2}{{\tan }^2}x}}} dx$

$= \int_0^{\pi /2} {\frac{{\frac{{\sin x}}{{\cos x}}}}{{1 + {m^2} \cdot \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}} dx$

$= \int_0^{\pi /2} {\frac{{\frac{{\sin x}}{{\cos x}}}}{{\frac{{{{\cos }^2}x + {m^2}{{\sin }^2}x}}{{{{\cos }^2}x}}}}} dx$

$= \int_0^{\pi /2} {\frac{{\sin x\cos xdx}}{{1 - {{\sin }^2}x + {m^2}{{\sin }^2}x}}} dx$

$= \int_0^{\pi /2} {\frac{{\sin x\cos x}}{{1 - {{\sin }^2}x\left( {1 - {m^2}} \right)}}} dx$

Let's put ${\sin ^2}x = t$

$\Rightarrow$ $2\sin x\cos xdx = dt$

therefore,$I = \frac{1}{2}\int_0^1 {\frac{{dt}}{{1 - t\left( {1 - {m^2}} \right)}}}$

$= \frac{1}{2}\left[ { - \log \left| {1 - t\left( {1 - {m^2}} \right)} \right| \cdot \frac{1}{{1 - {m^2}}}} \right]_0^1$

$= \frac{1}{2}\left[ { - \log \left| {1 - 1 + {m^2}} \right| \cdot \frac{1}{{1 + {m^2}}} + \log |1| \cdot \frac{1}{{1 - {m^2}}}} \right]$

$= \frac{1}{2}\left[ { - \log \left| {{m^2}} \right| \cdot \frac{1}{{1 - {m^2}}}} \right] = \frac{2}{2} \cdot \frac{{\log m}}{{\left( {{m^2} - 1} \right)}}$

$= \log \frac{m}{{{m^2} - 1}}$