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

$\int_{2}^{3} \frac{x}{x^{2}-1} d x=$

$\int_1^2 x \sqrt{4-x^2} \, dx =$

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

The value of $\int_1^{e^2} \frac{dx}{x(1 + \ln x)^2}$ is

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module