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

1 Answer

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

Explanation:

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