if

$f^{\prime}(x)=\tan ^{-1}(\sec x+\tan x)$

$=\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right)=\tan ^{-1}\left(\frac{1-\cos \left(\frac{\pi}{2}+x\right)}{\sin \left(\frac{\pi}{2}+x\right)}\right)$

$=\tan ^{-1}\left(\frac{2 \sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2 \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}\right)$

$=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right)=\frac{\pi}{4}+\frac{x}{2}$

Integrate both sides, we get

$\int\left(f^{\prime}(x)\right) d x=\int\left(\frac{\pi}{4}+\frac{x}{2}\right) d x$

$f(x)=\frac{\pi}{4} x+\frac{x^{2}}{4}+C$

$\because \quad f(0)=0$

$C=0 \Rightarrow f(x)=\frac{\pi}{4} x+\frac{x^{2}}{4}$

So, $\quad f(1)=\frac{\pi+1}{4}$