If int frac{sqrt{x}}{x(x+1)} dx = k tan^{-1} | vedclass

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(B) Let $I = \int \frac{\sqrt{x}}{x(x+1)} dx$. Substitute $x = t^2$,so $dx = 2t dt$. Then $I = \int \frac{t}{t^2(t^2+1)} (2t dt) = \int \frac{2t^2}{t^

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$k=2, m=\sqrt{x}$

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(B) Let $I = \int \frac{\sqrt{x}}{x(x+1)} dx$. Substitute $x = t^2$,so $dx = 2t dt$. Then $I = \int \frac{t}{t^2(t^2+1)} (2t dt) = \int \frac{2t^2}{t^2(t^2+1)} dt$. $I = \int \frac{2}{t^2+1} dt = 2 	an^{-1}(t) + c$. Since $t = \sqrt{x}$,we have $I = 2 	an^{-1}(\sqrt{x}) + c$. Comparing this with $k 	an^{-1} m + c$,we get $k=2$ and $m=\sqrt{x}$.

If $\int \frac{\sqrt{x}}{x(x+1)} dx = k \tan^{-1} m + c$,(where $c$ is the constant of integration),then:

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