Srinivasa B

You have started it right. let, u = ln(2x+1), du = 2.dx/(2x+1) dv = dx, v = x now the integral will become, x.ln(2x+1) - int(2x.dx/(2x+1)) by algebraic manupulation, it will turn out to be, x.ln(2x+1) - int(dx) + int(dx/(2x+1)) which will yield: x.ln(2x+1) - x + ln(2x+1) /2 x.ln(2x+1) - x + ln(2x+1) /2 Verify your result by differentiating again.

Ya, if y = lx(x), y = ln(ax) = ln(a) + ln(x) so, dy/dx = 0 + lim (ln(x + dx) - ln(x)) / dx dx -> 0 = lim ln(1+ dx/x) / dx dx -> 0 now, to get into e pow x form, put n = x/dx now, dy/dx = (1/x) * lim ( ln(1 + 1/n) ^ n) n -> inf so, dy/dx = 1/x. Yuppie

Hi Guys, I have a fundamental doubt on differential of y = f(x) = ln(ax) where, ln - natual logorithm to base e, a - constant x - independent variable. I need to find the value of dy/dx from first principles. Can anybody help me on this? Thanks, Srini