Integrating the Square Root of a Fraction: How to Solve This Tricky Integral

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Homework Statement



\int \sqrt {\frac{1+x}{1-x}} dx = ?

The solution is:

= \arcsin{x} -\sqrt{1-x^2} + const

If someone could help me to find the solution I'll be very pleased! :-)
Thanks everybody!
 
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Try multiplying the integrand by \frac{\sqrt{1+x}}{\sqrt{1+x}} and remember that (1+x)(1-x) = 1 - x^2. If you consider what the derivative of arcsin(x) is, it's easier to see what you should be trying to do to the integral. You may also need to do a substitution.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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