How do you find the integral of #dx/(x(sqrt(3 + x^2)))#?

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Answer 1 · 2018-07-04T21:44:54

#\int \frac{dx}{x\sqrt{3+x^2}}=1/\sqrt3\ln|\tan(1/2\tan^{-1}(x/\sqrt3))|+C#

Let #x=\sqrt3\tan\theta\implies dx=\sqrt3\sec^2\theta\ d\theta#

#\int \frac{dx}{x\sqrt{3+x^2}}#

#=\int \frac{\sqrt3\sec^2\theta\ d\theta}{\sqrt3\tan\theta\sqrt{3+3\tan^2\theta}}#

#=\int \frac{\sec^2\theta\ d\theta}{\sqrt3\tan\theta\sqrt{1+\tan^2\theta}}#

#=\int \frac{\sec^2\theta\ d\theta}{\sqrt3\tan\theta\sec\theta}#

#=1/\sqrt3\int \frac{\sec\theta\ d\theta}{\tan\theta}#

#=1/\sqrt3\int \frac{\sec\theta\cos\theta\ d\theta}{\sin\theta}#

#=1/\sqrt3\int \frac{ d\theta}{\sin\theta}#

#=1/\sqrt3\int \cosec \thetad\theta#

#=1/\sqrt3\ln|\tan(\theta/2)|+C#

#=1/\sqrt3\ln|\tan(1/2\tan^{-1}(x/\sqrt3))|+C#