What Substitution Solves This Integral?

sleepwalker27
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1.
http://www.imageurlhost.com/images/cnj1t05jh6e4fxqy4i5_integral.png
I know that this integral is solved by the sustitution method

The Attempt at a Solution


I tried converting the integral to the form of Arctanx, but that x2 on the numerator ruined everything. Thanks
 
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sleepwalker27 said:
1.
http://www.imageurlhost.com/images/cnj1t05jh6e4fxqy4i5_integral.png
I know that this integral is solved by the sustitution method

The Attempt at a Solution


I tried converting the integral to the form of Arctanx, but that x2 on the numerator ruined everything. Thanks
Please show us the substitution you used, which sounds like a trig substitution.
 
Mark44 said:
Please show us the substitution you used, which sounds like a trig substitution.
I tried to made the integral to the form ∫dx/x2+1 so that their solution is something like arctanx
 
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sleepwalker27 said:
I tried to made the integral to the form ∫dx/x2+1 so that their solution is something like arctanx
I figured you did something like that, but that isn't what I asked you for. Please show me your substitution.
 
Did you not notice that the numerator and denominator have the same degree?

\frac{2x^2}{2x^2+ 1}= 1- \frac{1}{2x^2+ 1}
 
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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