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

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(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

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

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(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 $B$.

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

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