int frac{d x}{x+sqrt{x-1}} = | vedclass

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(D) Given integral,$I = \int \frac{d x}{x+\sqrt{x-1}}$Put $x-1 = t^2 \Rightarrow dx = 2t \, dt$Then $I = \int \frac{2t}{(t^2+1)+t} \, dt = \int \frac{

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$\log _e|x+\sqrt{x-1}|-\frac{2}{\sqrt{3}} 	an ^{-1}\left(\frac{2 \sqrt{x-1}+1}{\sqrt{3}}ight)+c$

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(D) Given integral,$I = \int \frac{d x}{x+\sqrt{x-1}}$Put $x-1 = t^2 \Rightarrow dx = 2t \, dt$Then $I = \int \frac{2t}{(t^2+1)+t} \, dt = \int \frac{(2t+1)-1}{t^2+t+1} \, dt$$= \int \frac{2t+1}{t^2+t+1} \, dt - \int \frac{dt}{t^2+t+1}$$= \log _e|t^2+t+1| - \int \frac{dt}{(t+\frac{1}{2})^2 + \frac{3}{4}}$$= \log _e|t^2+t+1| - \frac{2}{\sqrt{3}} 	an ^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{3}}{2}}ight) + c$$= \log _e|t^2+t+1| - \frac{2}{\sqrt{3}} 	an ^{-1}\left(\frac{2t+1}{\sqrt{3}}ight) + c$On substituting $t = \sqrt{x-1}$,we get$I = \log _e|x+\sqrt{x-1}| - \frac{2}{\sqrt{3}} 	an ^{-1}\left(\frac{2\sqrt{x-1}+1}{\sqrt{3}}ight) + c$Hence,option $D$ is correct.

$\int \frac{d x}{x+\sqrt{x-1}} = $

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