Given:
[tex]\int ^1_{-1}\frac{1}{x^2+1}dx[/tex]
To evaluate:
As we know,
[tex]\int ^{}_{}\frac{1}{x^2+1^{}}dx=\tan ^{-1}(x)+C[/tex]
So, we have
[tex]\begin{gathered} \int ^1_{-1}\frac{1}{x^2+1}dx=\tan ^{-1}(1)-\tan ^{-1}(-1) \\ =\frac{\pi}{4}-(-\frac{\pi}{4}) \\ =\frac{\pi}{4}+\frac{\pi}{4} \\ =\frac{2\pi}{4} \\ =\frac{\pi}{2} \end{gathered}[/tex]
Hence, the answer is,
[tex]\frac{\pi}{2}[/tex]
