Question

The value of $\underset{-1}{\overset{0}{\int }}\frac{dx}{x^2+2x+2}$ is:

• A. $\frac{\pi }{4}$
• B. $0$
• C. $\pi$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $\frac{\pi }{4}$

Let $I=\underset{-1}{\overset{0}{\int }}\frac{dx}{x^2+2x+2}$
$=\underset{-1}{\overset{0}{\int }}\frac{dx}{(x+1{)}^2+1}$
$={\left[{\tan }^{-1}\right(x+1\left)\right]}_{-1}^0$
$={\tan }^{-1}1-{\tan }^{-1}0=\frac{\pi }{4}$