Question

${\displaystyle {\int }_{-2}^1\frac{dx}{x^2+4x+13}}$ = ______

Options

• ${\displaystyle \frac{\pi }{4}}$

• ${\displaystyle \frac{\pi }{6}}$

• ${\displaystyle \frac{\pi }{12}}$

• ${\displaystyle \frac{\pi }{3}}$

Answer 1

${\displaystyle {\int }_{-2}^1\frac{dx}{x^2+4x+13}}$ = ${\displaystyle \underline{\frac{\pi }{12}}}$

Explanation:

${\displaystyle {\int }_{-2}^1\frac{dx}{x^2+4x+13}}$ = ${\displaystyle {\int }_{-2}^1\frac{dx}{x^2+4x+4+9}}$

= ${\displaystyle {\int }_{-2}^1\frac{dx}{{(x+2)}^2+{\left(3\right)}^2}=\frac{1}{3}{\left[{\tan }^{-1}\left(\frac{x+2}{3}\right)\right]}_{-2}^1}$

= ${\displaystyle \frac{1}{3}[{\tan }^{-1}\left(1\right)-{\tan }^{-1}\left(0\right)]}$

= ${\displaystyle \frac{1}{3}(\frac{\pi }{4}-0)}$

= ${\displaystyle \frac{\pi }{12}}$