Question

$\underset{-1}{\overset{1}{\int }}\frac{1}{x^2+2x+5}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_{-1}^1\frac{1}{x^2+2x+5}\mathit{d}x\mathit{.}\mathrm{Then}, \\ \mathrm{I}={\int }_{-1}^1\frac{1}{\left({\mathrm{x}}^2+2\mathrm{x}+1\right)+4}\mathit{d}\mathrm{x} \\ \Rightarrow I={\int }_{-1}^1\frac{1}{{\left(x+1\right)}^2+2^2}\mathit{}\mathit{d}x \\ \Rightarrow I=\frac{1}{2}{\left[{\tan }^{-1}\frac{\left(x+1\right)}{2}\right]}_{-1}^1 \\ \Rightarrow I=\frac{1}{2}\left({\tan }^{-1}1-{\tan }^{-1}0\right) \\ \Rightarrow I=\frac{1}{2}\left(\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow I=\frac{\mathrm{\pi }}{8} \end{gathered}$$