int frac{ln |x|}{xsqrt{1 + ln |x|}} , dx equa | vedclass

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Question

(A) Let $t = 1 + \ln |x|$. Then $dt = \frac{1}{x} \, dx$,and $\ln |x| = t - 1$.Substituting these into the integral:$\int \frac{t - 1}{\sqrt{t}} \, dt

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$\frac{2}{3} \sqrt{1 + \ln |x|} (\ln |x| - 2) + c$

Vedclass · Answer

(A) Let $t = 1 + \ln |x|$. Then $dt = \frac{1}{x} \, dx$,and $\ln |x| = t - 1$.Substituting these into the integral:$\int \frac{t - 1}{\sqrt{t}} \, dt = \int (\sqrt{t} - \frac{1}{\sqrt{t}}) \, dt$$= \int (t^{1/2} - t^{-1/2}) \, dt$$= \frac{t^{3/2}}{3/2} - \frac{t^{1/2}}{1/2} + c$$= \frac{2}{3} t^{3/2} - 2 t^{1/2} + c$$= \frac{2}{3} \sqrt{t} (t - 3) + c$Substituting $t = 1 + \ln |x|$ back:$= \frac{2}{3} \sqrt{1 + \ln |x|} (1 + \ln |x| - 3) + c$$= \frac{2}{3} \sqrt{1 + \ln |x|} (\ln |x| - 2) + c$.

$\int \frac{\ln |x|}{x\sqrt{1 + \ln |x|}} \, dx$ equals :

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