How would you evaluate the integral of (2LnX)/X on the interval of 1 to 2?

Answer 1

Make the substitution u = ln(x). Then du = (1/x)dx

∫2*ln(x) * dx/x = ∫2*udu =2* (u^2)/2=u^2. Put back u=ln(x) to get the result l[n(x)]^2. The integral value is then [ln(2)]^2 - [ln(1)]^2 = [ln(2)]^2