Evaluate $\int\frac{1}{x\sqrt{1-x^2}}dx$ without using trigonometric substitution. The integral is
$$\int\frac{1}{x\sqrt{1-x^2}}dx\tag{1}$$
I tried solving it by parts, but that didn't work out. I couldn't integrate the result of substituting $t=1-x^2$ either.
The answer is
$$\ln\left|\dfrac{1-\sqrt{1-x^2}}{x}\right|$$
 A: Let $\sqrt{-x^{2}+1}=xt+1$ to transform the integral into an easier rational function. Rearrange for $x$ and we have $$x = \frac{2t}{-1-t^{2}}$$
Find the derivative $\frac{dx}{dt}$, then substitute $x$ and $dx$ into the integral and work towards your answer.
Solution:

 $$\frac{dx}{dt}=\frac{2t^{2}-2}{\left(-1-t^{2}\right)^{2}}$$


Our integral becomes thus $$\int\frac{\frac{2t^{2}-2}{\left(-1-t^{2}\right)^{2}}}{\frac{2t}{-1-t^{2}}\left(\frac{2t}{-1-t^{2}}+1\right)}dt$$ which simplifies nicely to $$\int\frac{1}{t}dt\ =\ \ln\left|t\right|\ + C$$ where $t=\frac{\sqrt{1-x^{2}}-1}{x}$

A: With the substitution $t=\sqrt{1-x^2}$, the integral can be integrated as follows
$$\int\frac{1}{x\sqrt{1-x^2}}dx=-\int \frac1{1-t^2}dt= \frac12 \ln\frac{1-t}{1+t}= \frac12\ln \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}+C
$$
which is the same as $\ln\frac{1-\sqrt{1-x^2}}{x}$ after rationalizing the denominator.
A: Edit: I just realised that you wanted to evaluate this integral without trigonometric substitution. I'm sorry if this answer is not of any use to you. I will keep it up for the benefit of other readers.

Because of the identity $\sin^2\theta+\cos^2\theta=1$, a good candidate for a substitution is $x=\sin\theta$ (with $-\pi/2\le\theta\le\pi/2$). If we rearrange $\cos^2\theta+\sin^2\theta=1$, we get
$$
\sqrt{\cos^2\theta} = |\cos\theta|=\sqrt{1-\sin^2\theta}
$$
but since $\cos\theta$ is nonnegative for $\theta\in[-\pi/2,\pi/2]$, this simplifies to
$$
\cos\theta=\sqrt{1-\sin^2\theta}
$$
Let us try using this substitution for the integral at hand. If $x=\sin\theta$, then $dx=\cos\theta \, d\theta$, and so
\begin{align}
\int \frac{dx}{x\sqrt{1-x^2}} &= \int \frac{\cos\theta \, d\theta}{\sin\theta\cos\theta} \\[5pt]
&= \int \csc\theta \\[5pt]
&= -\log|\csc\theta+\cot\theta| + C \label{*}\tag{*} \\[5pt]
&= -\log\left|\frac{1+\cos\theta}{\sin\theta}\right| + C \\[5pt]
&= -\log\left|\frac{1+\sqrt{1-x^2}}{x}\right| + C
\end{align}
Using the identity $-\log|y|=\log|y^{-1}|$, this becomes
$$
\int \frac{dx}{x\sqrt{1-x^2}} = \log\left|\frac{1-\sqrt{1-x^2}}{x}\right| + C
$$

If you are not familiar with how to integrate $\csc\theta$ as in $\eqref{*}$, then note that
$$
\int \csc\theta \, d\theta = \int\frac{\sin\theta}{1-\cos^2\theta} \, d\theta
$$
Try making the substitution $u=\cos\theta$ and be prepared to simplify your answer a lot!
