$\text{ If }{f}^{\text{'}}\left(x\right)={\tan }^{-1}(\sec x+\tan x),-\frac{\pi }{2}<x<\frac{\pi }{2},\text{ and }f\left(0\right)=0,\text{ then }f\left(1\right)\text{ is equal to: }$

$f^{\text{'}}\left(x\right)={\tan }^{-1}(\sec x+\tan x)$

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

          $$\begin{gathered} ={\tan }^{-1}\left(\tan \left(\frac{\pi }{4}+\frac{x}{2}\right)\right)=\frac{\pi }{4}+\frac{x}{2} \\ \int f^{\text{'}}\left(x\right)dx=\int \left(\frac{\pi }{4}+\frac{x}{2}\right)dx \\ f\left(x\right)=\frac{\pi }{4}x+\frac{x^2}{4}+c \end{gathered}$$

$\begin{array}{l}f\left(0\right)=c=0\Rightarrow f\left(x\right)=\frac{\pi }{4}x+\frac{x^2}{4}\\ \text{ So }f\left(1\right)=\frac{\pi +1}{4}\end{array}$