int_0^4 frac{1}{1+sqrt{x}} , dx = dots | vedclass

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(C) Let $I = \int_0^4 \frac{1}{1+\sqrt{x}} \, dx$. Substitute $x = t^2$,so $dx = 2t \, dt$. When $x = 0, t = 0$ and when $x = 4, t = 2$. Then,$I = \in

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $I = \int_0^4 \frac{1}{1+\sqrt{x}} \, dx$. Substitute $x = t^2$,so $dx = 2t \, dt$. When $x = 0, t = 0$ and when $x = 4, t = 2$. Then,$I = \int_0^2 \frac{2t}{1+t} \, dt$. $I = 2 \int_0^2 \frac{(1+t)-1}{1+t} \, dt$. $I = 2 \int_0^2 \left(1 - \frac{1}{1+t}\right) \, dt$. $I = 2 [t - \ln(1+t)]_0^2$. $I = 2 [(2 - \ln 3) - (0 - \ln 1)]$. $I = 2(2 - \ln 3) = 4 - 2\ln 3 = 4 - \ln(3^2) = 4 - \ln 9$. Since $4 = \ln(e^4)$,we have $I = \ln(e^4) - \ln 9 = \ln \left(\frac{e^4}{9}\right)$.

$\int_0^4 \frac{1}{1+\sqrt{x}} \, dx = \dots$

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