int_{-1}^{0} frac{dx}{x^2 + 2x + 2} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int_{-1}^{0} \frac{dx}{x^2 + 2x + 2}$.First,complete the square in the denominator: $x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1^2

Vedclass · Accepted Answer

$\pi / 4$

Vedclass · Answer

(B) Let $I = \int_{-1}^{0} \frac{dx}{x^2 + 2x + 2}$.First,complete the square in the denominator: $x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1^2$.Thus,the integral becomes $I = \int_{-1}^{0} \frac{dx}{(x + 1)^2 + 1^2}$.Using the standard integral formula $\int \frac{dx}{x^2 + a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get:$I = [	an^{-1}(x + 1)]_{-1}^{0}$.Evaluating the limits:$I = 	an^{-1}(0 + 1) - 	an^{-1}(-1 + 1) = 	an^{-1}(1) - 	an^{-1}(0)$.Since $	an^{-1}(1) = \frac{\pi}{4}$ and $	an^{-1}(0) = 0$,we have:$I = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

$\int_{-1}^{0} \frac{dx}{x^2 + 2x + 2} = $

Similar Questions

$\int_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\,=$

$\int_0^{\frac{\pi}{4}} \frac{\sec x}{1+2 \sin ^2 x} d x=$

Evaluate the definite integral: $\int_{0}^{\frac{\pi}{4}}\left(2 \sec ^{2} x+x^{3}+2\right) d x$

$\int_{0}^{2\pi} |\sin x| \, dx = $

$\int_{-2}^3 |1-x^2| dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module