Integrating $\int \frac{1}{x\sqrt{3-x^2}}dx$ without trig sub

Question

So I am evaluating $\int \frac{1}{x\sqrt{3-x^2}}dx$ without using trig sub integrals. So far I have

$$u=\sqrt{3-x^2}, x^2=3-u^2,du=-\frac{x}{\sqrt{3-x^2}}dx, dx = -\frac{\sqrt{3-x^2}}{x}$$

So rewriting I get

$$\int \frac{1}{x\sqrt{3-x^2}}dx=-\int\frac{1}{3-u^2}du$$

Then I use partial fraction decamp to find that both variables are $\frac{1}{2\sqrt{3}}$ so I get

$$-\int\frac{1}{3-u^2}du=-\frac{1}{2\sqrt{3}}\bigg[\int \frac{1}{\sqrt{3}+u}du+\frac{1}{\sqrt{3}-u}du\bigg]=-\frac{1}{2\sqrt{3}}\bigg( \ln \bigg\vert \sqrt{3} +\sqrt{3-x^2}\bigg \vert - \ln \bigg \vert \sqrt{3}-\sqrt{3-x^2} \bigg \vert \bigg)+C$$

But this isn't the answer, where am I going wrong?

Answer 1

Your answer looks fine to me. To double check, you can differentiate your answer, or confer with wolfram, which gives

$$\int \frac{1}{x\sqrt {3-x^2}}=-\frac{1}{\sqrt 3}\tanh^{-1}\sqrt{1-x^2/3}+C,$$

and using the identity

$$\tanh^{-1}t=\frac{1}{2}\ln \left(\frac{1+t}{1-t}\right),t\in(-1,1)$$

we have $$\begin{align}-\frac{1}{\sqrt 3}\tanh^{-1}\sqrt{1-x^2/3}+C&=-\frac{1}{2\sqrt 3}\ln \left(\frac{1+\sqrt{1-x^2/3}}{1-\sqrt{1-x^2/3}}\right)+C\\ &=-\frac{1}{2\sqrt 3}\ln \left(\frac{\sqrt 3+\sqrt{3-x^2}}{\sqrt 3-\sqrt{3-x^2}}\right)+C\\ &=-\frac{1}{2\sqrt 3}\left(\ln \left(\sqrt 3+\sqrt{3-x^2}\right)-\ln \left(\sqrt 3-\sqrt{3-x^2}\right)\right)+C \end{align}.$$