How Does the Integral of 1/(4y-1) with Respect to x Evaluate from 0 to 1?

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integral of 1/(4y-1) dx from 0 to 1

im stumped
 
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nick727kcin said:
integral of 1/(4y-1) dx from 0 to 1

im stumped

Whats the integral of 1/y ?

Ohh I missed what LocationX pointed out, are you sure it's a function of y that you are integrating with respect to x? Or is the y supposed to be x, or the dx dy?
 
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if it is with respects to x, then the integral is just \frac{x}{4y-1}
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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