Daniel Ward	Posted on Tuesday, 06 May, 2003 - 09:57 pm: Use the substitution x=1/u to find the value of ${\displaystyle {\int }_{1/2}^2\frac{\log x}{1+x^2}\hspace{0.5em}\backslash dx}$ Which is all fine and dandy, except when you do this, I make it that you get back to exactly the same function, only of u instead of x. What do I do ???
David Loeffler	Posted on Tuesday, 06 May, 2003 - 10:13 pm: Hmm. When you do the change of variables, three minus signs appear (not two) - one from the log, one from the fact that du/dx is negative, and a third from the limits having switched around. So your integral ${\displaystyle ={\int }_2^{1/2}\frac{-\log u}{1+u^{-2}}.\frac{-1}{u^2}\hspace{0.5em}\mathrm{du}}$ ${\displaystyle =-{\int }_{1/2}^2\frac{\log u}{1+u^2}\hspace{0.5em}\mathrm{du}}$ This is minus the original integral, so the integral is zero. David
Daniel Ward	Posted on Thursday, 08 May, 2003 - 07:56 pm: aaah. Thank you.