Integrate the function frac{1}{x-sqrt{x}}. | vedclass

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Question

(A) The given integral is $I = \int \frac{1}{x-\sqrt{x}} dx$. We can rewrite the denominator as $\sqrt{x}(\sqrt{x}-1)$. So,$I = \int \frac{1}{\sqrt{x}

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

Vedclass · Answer

(A) The given integral is $I = \int \frac{1}{x-\sqrt{x}} dx$. We can rewrite the denominator as $\sqrt{x}(\sqrt{x}-1)$. So,$I = \int \frac{1}{\sqrt{x}(\sqrt{x}-1)} dx$. Let $t = \sqrt{x}-1$. Then,$dt = \frac{1}{2\sqrt{x}} dx$,which implies $\frac{dx}{\sqrt{x}} = 2 dt$. Substituting these into the integral,we get $I = \int \frac{2}{t} dt$. Integrating with respect to $t$,we obtain $I = 2 \log |t| + C$. Substituting back $t = \sqrt{x}-1$,we get $I = 2 \log |\sqrt{x}-1| + C$,where $C$ is an arbitrary constant.

Integrate the function $\frac{1}{x-\sqrt{x}}$.

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