Solution to Exercise #29 (Section 8-4 of Calculus, Anton, Bivens, Davis)Prepared by Douglas B. Meade30 Sep 2008restart;with( Student[Calculus1] ):Formulate the arclength of y=ln(x), 1 <= x <= 2, as a definite integral.L := Int( sqrt(1+x^2)/x, x=1..2 );We can just ask Maple for the answer,value( L );but arctanh is not how we would express the answer. We can ask Maple to convert this to exponentials and logarithms:convert( , expln );This looks better, but a real mess. The numerical approximation to this length isevalf( );Now, let's worry about how this definite integral can be found. Let's see what techniques Maple thinks could be useful on this problem.Hint( L );Three changes of variables. Of these, the first looks like the trig substitution we expect.Rule[%[1]](L);From here, the hints take us through the same steps that we did in class.Hint(%);Rule[%]();Hint(%);Rule[%]();Hint(%);Rule[%]();Hint(%);Rule[%]();Hint(%);Rule[%]();Hint(%);Rule[%]();Hint(%);Rule[%]();Hint(%);Rule[%]();Hint(%);Rule[%]();Hint(%);Rule[%]();Hint(%);Rule[%]();Success! This looks very much like what we found in class.Recall that at the first step Maple gave us three different changes of variables. Why don't you explore one or both of these. Do you get to the same result? Is either approach easier or more difficult than what we did?