Given, $\mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos ^2 x} d x$
$\mathrm{I}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{2-\cos ^2 x} d x+\frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos ^2 x} d x$
Let, $I_1=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x}{2-\cos ^2 x} d x$
$f(x)=\frac{x}{2-\cos ^2 x}$
$f(-x)=\frac{-x}{2-\cos 2(-x)}=\frac{-x}{2-\cos 2 x}=-f(x)$
$f(x)$ is an odd function.
Thus, $I_1=0$
Let $I_1=\frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2-\cos 2 x} d x$
$g(x)=\frac{\pi}{4(2-\cos 2 x)}$
$g(-x)=\frac{\pi}{4(2-\cos 2(-x))}=\frac{\pi}{4(2-\cos 2 x)}=g(x)$
$\therefore \quad g(x)$ is an even function
Thus, $\mathrm{I}_2=2 \times \frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{d x}{2-\cos 2 x}$
Now, $\mathrm{I}=\mathrm{I}_2=\frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{d x}{2-\frac{1-\tan ^2 x}{1+\tan ^2 x}}$
$\begin{aligned}
& \mathrm{I}=\frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{1+\tan ^2 x}{2+\tan ^2 x-1+\tan ^2 x} d x \\
& \mathrm{I}=\frac{\pi}{4} \int_0^{\frac{\pi}{4}} \frac{\sec ^2 x}{1+3 \tan ^2 x} d x
\end{aligned}$
Put $\tan x=t \Rightarrow \sec ^2 x d x=d t$
Limits : $x=0 \quad \Rightarrow \mathrm{t}=0$
$\begin{aligned}
& x=\frac{\pi}{4} \quad \Rightarrow \mathrm{t}=1 \\
& \therefore \quad \mathrm{I}=\frac{\pi}{4} \int_0^1 \frac{d t}{1+3 t^2} \\
& \mathrm{I}=\frac{\pi}{2} \times \frac{1}{\sqrt{3}}\left[\tan ^{-1} \sqrt{3} t\right]_0^1 \\
& \mathrm{I}=\frac{\pi}{2 \sqrt{3}}\left[\tan ^{-1}(\sqrt{3})-\tan ^{-1}(0)\right] \\
& \mathrm{I}=\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}\right)=\frac{\pi^2}{6 \sqrt{3}}
\end{aligned}$