int frac{dx}{1+sqrt{x}} = | vedclass

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Question

(A) Let $I = \int \frac{dx}{1+\sqrt{x}}$.Substitute $\sqrt{x} = t$,which implies $x = t^2$,so $dx = 2t \, dt$.Substituting these into the integral:$I

Vedclass · Accepted Answer

$2\sqrt{x} - 2\log |1+\sqrt{x}| + c$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{1+\sqrt{x}}$.Substitute $\sqrt{x} = t$,which implies $x = t^2$,so $dx = 2t \, dt$.Substituting these into the integral:$I = \int \frac{2t}{1+t} \, dt$.$I = 2 \int \frac{t}{1+t} \, dt$.$I = 2 \int \frac{(t+1) - 1}{1+t} \, dt$.$I = 2 \int \left( 1 - \frac{1}{1+t} \right) \, dt$.$I = 2 \left( \int 1 \, dt - \int \frac{1}{1+t} \, dt \right)$.$I = 2(t - \log|1+t|) + c$.Substituting $t = \sqrt{x}$ back:$I = 2\sqrt{x} - 2\log|1+\sqrt{x}| + c$.

$\int \frac{dx}{1+\sqrt{x}} = $

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