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

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

Vedclass · Accepted Answer

$2(\sqrt{x} - 	an^{-1} \sqrt{x}) + c$, where $c$ is a constant of integration.

Vedclass · Answer

(B) Let $I = \int \frac{\sqrt{x}}{x+1} \,dx$. Substitute $\sqrt{x} = t$, which implies $x = t^2$ and $dx = 2t \,dt$. Substituting these into the integral: $I = \int \frac{t}{t^2+1} (2t) \,dt = \int \frac{2t^2}{t^2+1} \,dt$. Rewrite the numerator: $I = 2 \int \frac{t^2+1-1}{t^2+1} \,dt = 2 \int \left(1 - \frac{1}{t^2+1}ight) \,dt$. Integrating term by term: $I = 2 \left( \int 1 \,dt - \int \frac{1}{t^2+1} \,dt ight) = 2(t - 	an^{-1} t) + c$. Substituting $t = \sqrt{x}$ back: $I = 2(\sqrt{x} - 	an^{-1} \sqrt{x}) + c$.

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

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