int_0^1 frac{1}{2+sqrt{x}} , dx = | vedclass

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

Question

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

Vedclass · Accepted Answer

$2 \log \left(\frac{4 e}{9}\right)$

Vedclass · Answer

(B) Let $I = \int_0^1 \frac{1}{2+\sqrt{x}} \, dx$. Substitute $\sqrt{x} = t$,so $x = t^2$ and $dx = 2t \, dt$. When $x = 0$,$t = 0$. When $x = 1$,$t = 1$. Thus,$I = \int_0^1 \frac{2t}{2+t} \, dt$. $I = 2 \int_0^1 \frac{t+2-2}{t+2} \, dt = 2 \int_0^1 \left(1 - \frac{2}{t+2}\right) \, dt$. $I = 2 [t - 2 \log |t+2|]_0^1$. $I = 2 [(1 - 2 \log 3) - (0 - 2 \log 2)]$. $I = 2 [1 - 2 \log 3 + 2 \log 2] = 2 [1 + 2 \log(2/3)]$. $I = 2 [\log e + \log(4/9)] = 2 \log(4e/9)$. Therefore,the correct option is $B$.

$\int_0^1 \frac{1}{2+\sqrt{x}} \, dx =$

Similar Questions

The value of $\int_0^2 \frac{3^{\sqrt{x}}}{\sqrt{x}} \, dx$ is

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

$\int_0^{\pi /2} \frac{\sin x \cos x}{1 + \sin^4 x} \, dx = $

$\int_0^{\pi /6} \frac{\sin x}{\cos^3 x} \, dx = $

$\int_1^e \frac{1 + \log x}{x} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module