Integrating Logarithmic Functions: Understanding the Integral of Logy

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Homework Statement



\int\int-xlogy dy dx

the only problem i have is finding the integral of logy
i know i have to integrate my parts with u = logy and dv=dy
the book says the the integral of logy is ylogy-y but isn't the derivative of logy equal to 1/ylna where a is the base?
so isn't the integral of logy equal to ylogy-y/ln10 because of the base 10?
can someone clarify this for me?
 
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I know some calculus books (especially the older ones I have looked at) use log(y) to designate the natural logarithm. Are you certain this isn't the case?
 
oooo it is thanks makes a lot more sense now
 

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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