Question

Evaluate the integrals using substitution.
${\int }_{-1}^1\frac{1}{x^2+2x+5}dx.$

Answer 1

${\int }_{-1}^1\frac{1}{x^2+2x+5}dx={\int }_{-1}^1\frac{1}{(x^2+2x+1)+4}dx={\int }_{-1}^1\frac{1}{(x+1{)}^2+2^2}dx=\frac{1}{2}{\left[ta{n}^{-1}\frac{x+1}{2}\right]}_{-1}^1[\because \int \frac{dx}{a+x^2}=\frac{1}{a}ta{n}^{-1}\frac{x}{a}]=\frac{1}{2}\left[ta{n}^{-1}\right(\frac{2}{2})-ta{n}^{-1}(\frac{0}{2}\left)\right]=\frac{1}{2}\left[ta{n}^{-1}1\right]=\frac{1}{2}\times \frac{\pi }{4}=\frac{\pi }{8}.$