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

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Question

(A) To evaluate the integral $I = \int \frac{\sqrt{x}}{1+x} dx$,substitute $x = t^2$,so $dx = 2t dt$.Substituting these into the integral:$I = \int \f

Vedclass · Accepted Answer

$2(\sqrt{x} - 	an^{-1}\sqrt{x}) + c$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{\sqrt{x}}{1+x} dx$,substitute $x = t^2$,so $dx = 2t dt$.Substituting these into the integral:$I = \int \frac{t}{1+t^2} (2t dt) = 2 \int \frac{t^2}{1+t^2} dt$$I = 2 \int \frac{t^2+1-1}{1+t^2} dt = 2 \int \left(1 - \frac{1}{1+t^2}ight) dt$$I = 2 \left( \int 1 dt - \int \frac{1}{1+t^2} dt ight)$$I = 2(t - 	an^{-1}t) + c$Substituting $t = \sqrt{x}$ back:$I = 2(\sqrt{x} - 	an^{-1}\sqrt{x}) + c$.

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

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