<s>[INST] 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
$$\ln\left|\dfrac{1-\sqrt{1-x^2}}{x}\right|$$
 [/INST] 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.
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