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Mathematica gives $$g(a)=\frac{\pi}{2} \, \text{sgn}(a) \ln(1+|a|) $$ for real $a$.

This can be obtained as follows: $g(0)=0$ and $$g'(a)=\int_0^{\pi/2}\frac{dx}{1+a^2\tan^2 x}=\int_0^\infty\frac{du}{(1+a^2u^2)(1+u^2)}=\frac{\pi/2}{1+|a|}, $$ where we used the substitution $u=\tan x$; the latter integral can be taken by partial fraction decomposition: $$\frac1{(1+a^2u^2)(1+u^2)}=\frac{1}{\left(1-a^2\right) \left(1+u^2\right)} - \frac{a^2}{(1-a^2) \left(1+a^2 u^2\right)} $$ for real $a$ with $|a|\ne1$; the case $|a|=1$ can be obtained by continuity.

The answer is $$g(a)=\frac{\pi}{2} \, \text{sgn}(a) \ln(1+|a|) $$ for real $a$.

This can be obtained as follows: $g(0)=0$ and $$g'(a)=\int_0^{\pi/2}\frac{dx}{1+a^2\tan^2 x}=\int_0^\infty\frac{du}{(1+a^2u^2)(1+u^2)}=\frac{\pi/2}{1+|a|}, $$ where we used the substitution $u=\tan x$; the latter integral can be taken by partial fraction decomposition: $$\frac1{(1+a^2u^2)(1+u^2)}=\frac{1}{\left(1-a^2\right) \left(1+u^2\right)} - \frac{a^2}{(1-a^2) \left(1+a^2 u^2\right)} $$ for real $a$ with $|a|\ne1$; the case $|a|=1$ can be obtained by continuity.

Mathematica gives $$g(a)=\frac{\pi}{2} \, \text{sgn}(a) \ln(1+|a|) $$ for real $a$.

This can be obtained as follows: $g(0)=0$ and $$g'(a)=\int_0^{\pi/2}\frac{dx}{1+a^2\tan^2 x}=\int_0^\infty\frac{du}{(1+a^2u^2)(1+u^2)}=\frac{\pi/2}{1+|a|}, $$ where we used the substitution $u=\tan x$; the latter integral can be taken by partial fraction decomposition.

Mathematica gives $$g(a)=\frac{\pi}{2} \, \text{sgn}(a) \ln(1+|a|) $$ for real $a$.

This can be obtained as follows: $g(0)=0$ and $$g'(a)=\int_0^{\pi/2}\frac{dx}{1+a^2\tan^2 x}=\int_0^\infty\frac{du}{(1+a^2u^2)(1+u^2)}=\frac{\pi/2}{1+|a|}, $$ where we used the substitution $u=\tan x$; the latter integral can be taken by partial fraction decomposition: $$\frac1{(1+a^2u^2)(1+u^2)}=\frac{1}{\left(1-a^2\right) \left(1+u^2\right)} - \frac{a^2}{(1-a^2) \left(1+a^2 u^2\right)} $$ for real $a$ with $|a|\ne1$; the case $|a|=1$ can be obtained by continuity.

Mathematica gives $$g(a)=\frac{\pi}{2} \, \text{sgn}(a) \ln(1+|a|) $$ for real $a$.

This can be obtained as follows: $g(0)=0$ and $$g'(a)=\int_0^{\pi/2}\frac{dx}{1+a^2\tan^2 x}=\int_0^\infty\frac{du}{(1+a^2u^2)(1+u^2)}=\frac\pi{2(1+|a|)}, $$ where we used the substitution $u=\tan x$; the latter integral can be taken by partial fraction decomposition.

Mathematica gives $$g(a)=\frac{\pi}{2} \, \text{sgn}(a) \ln(1+|a|) $$ for real $a$.

This can be obtained as follows: $g(0)=0$ and $$g'(a)=\int_0^{\pi/2}\frac{dx}{1+a^2\tan^2 x}=\int_0^\infty\frac{du}{(1+a^2u^2)(1+u^2)}=\frac{\pi/2}{1+|a|}, $$ where we used the substitution $u=\tan x$; the latter integral can be taken by partial fraction decomposition.