How was the integral of x^{-1/2}*(1-x)^-1 derived?

  • Thread starter Thread starter imurme8
  • Start date Start date
  • Tags Tags
    Integral
imurme8
Messages
45
Reaction score
0
Claim: \int\frac{dx}{\sqrt{x}(1-x)}=\log{\frac{1+\sqrt{x}}{1-\sqrt{x}}}
Derivation confirms this, but how was this answer arrived at? IBP seems not to work, can't find a good u-substitution...
 
Physics news on Phys.org
What about the substitution x=u^2, followed by partial fraction decomposition?
 
Try writing:

\frac{1}{\sqrt{x}(1-x)} = \frac{1}{\sqrt{x}(1+\sqrt{x})(1-\sqrt{x})}

then use a u substitution and partial fractions
 
Thanks guys!
 
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...
Back
Top