If \(\mathrm{f}^{\prime}(\mathrm{x})=\tan ^{-1}(\sec \mathrm{x}+ \tan \mathrm{x}), \frac{-\pi}{2}<\mathrm{x}<\frac{\pi}{2}\) and \(\mathrm{f}(0)=0\), then \(\mathrm{f}(1)=\)

\(\mathrm{f}^{\prime}(\mathrm{x})=\tan ^{-1}(\sec \mathrm{x}+\tan \mathrm{x})=\tan ^{-1}\left(\frac{1+\sin \mathrm{x}}{\cos \mathrm{x}}\right) \)
\( =\tan ^{-1}\left(\frac{\cos \frac{\mathrm{x}}{2}+\sin \frac{\mathrm{x}}{2}}{\cos \frac{\mathrm{x}}{2}-\sin \frac{\mathrm{x}}{2}}\right)=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\right) \)
\( =\frac{\pi}{4}+\frac{\mathrm{x}}{2}\)
Now \(f(x)=\int f^{\prime}(x) d x=\int\left(\frac{\pi}{4}+\frac{x}{2}\right) d x=\frac{\pi}{4} x+\frac{x^2}{4}+C\)
\(
\because \mathrm{f}(0)=0 \Rightarrow \mathrm{c}=0
\)
Hence, \(f(x)=\frac{x^2+\pi x}{4}\)
\(
\Rightarrow \mathrm{f}(1)=\frac{1^2+\pi \times 1}{4}=\frac{\pi+1}{4}
\)