<s>[INST] Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$ $$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$
How to integrate?
I tried the substitution $x=\sin\theta$ but didn't work.
 [/INST] Let $ x = \sin \theta$
$$
\int \frac{1}{\sin \theta} \cdot \sqrt \frac{1-\sin \theta}{1 + \sin \theta} \cdot \cos \theta \; d \theta \;\;= \;\; \int \frac{\cos\theta}{\sin\theta}\cdot \sqrt\frac{(1-\sin\theta)^2}{(1+\sin\theta)(1-\sin\theta)} \; d\theta
\\
= \;\; \int \frac{\cos\theta}{\sin\theta}\cdot \frac{1-\sin\theta}{\cos\theta}\; d\theta \;\; = \;\; \int (\csc\theta \; - 1) \; d\theta
$$
Then convert back in terms of x
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