If int frac{1}{(1 + x)sqrt{x}} , dx = f(x) + | vedclass

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(B) Let $I = \int \frac{1}{(1 + x)\sqrt{x}} \, dx$. We can rewrite the integral as $I = \int \frac{1}{(1 + (\sqrt{x})^2)\sqrt{x}} \, dx$. Let $\sqrt{x

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$2	an^{-1}\sqrt{x}$

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(B) Let $I = \int \frac{1}{(1 + x)\sqrt{x}} \, dx$. We can rewrite the integral as $I = \int \frac{1}{(1 + (\sqrt{x})^2)\sqrt{x}} \, dx$. Let $\sqrt{x} = t$. Then,differentiating both sides with respect to $x$,we get $\frac{1}{2\sqrt{x}} \, dx = dt$,which implies $\frac{dx}{\sqrt{x}} = 2 \, dt$. Substituting these into the integral,we get $I = \int \frac{2 \, dt}{1 + t^2}$. Integrating,we obtain $I = 2	an^{-1}(t) + A$. Substituting $t = \sqrt{x}$ back,we get $I = 2	an^{-1}(\sqrt{x}) + A$. Comparing this with $f(x) + A$,we find $f(x) = 2	an^{-1}(\sqrt{x})$.

If $\int \frac{1}{(1 + x)\sqrt{x}} \, dx = f(x) + A$,where $A$ is any arbitrary constant,then the function $f(x)$ is

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