int_0^3 frac{dx}{(x+2) sqrt{x+1}} = | vedclass

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

Question

(B) Let $I = \int_0^3 \frac{dx}{(x+2) \sqrt{x+1}}$.Substitute $t = \sqrt{x+1}$,so $t^2 = x+1$ and $x = t^2 - 1$.Then $dx = 2t \, dt$.When $x = 0$,$t =

Vedclass · Accepted Answer

$2 	an^{-1}\left(\frac{1}{3}ight)$

Vedclass · Answer

(B) Let $I = \int_0^3 \frac{dx}{(x+2) \sqrt{x+1}}$.Substitute $t = \sqrt{x+1}$,so $t^2 = x+1$ and $x = t^2 - 1$.Then $dx = 2t \, dt$.When $x = 0$,$t = 1$. When $x = 3$,$t = 2$.The integral becomes $I = \int_1^2 \frac{2t \, dt}{(t^2 - 1 + 2) t} = \int_1^2 \frac{2 \, dt}{t^2 + 1}$.Integrating,we get $I = [2 	an^{-1}(t)]_1^2 = 2(	an^{-1}(2) - 	an^{-1}(1))$.Using the formula $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,$I = 2 	an^{-1}\left(\frac{2-1}{1+2(1)}ight) = 2 	an^{-1}\left(\frac{1}{3}ight)$.

$\int_0^3 \frac{dx}{(x+2) \sqrt{x+1}} = $

Similar Questions

Evaluate $\int_{\sin \theta}^{\cos \theta} f(x \tan \theta) \, dx$,where $\theta \neq \frac{n \pi}{2}, n \in I$.

$\int_1^{e^2} \frac{dx}{x(1+\log x)^2} = $

The following integral $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(2 \operatorname{cosec} x)^{17} d x$ is equal to

The value of the definite integral,$\int\limits_0^{\sqrt {\ln \left( {\frac{\pi }{2}} \right)} } {\cos \left( {{e^{{x^2}}}} \right)} \cdot 2x {e^{{x^2}}}dx$ is

$\int_{\pi /4}^{\pi /2} \cos \theta \csc^2 \theta \, d\theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module