Use the substitution u=2+ln(t) to find the exact value of the antiderivative of 1/(t(2+ln(t))^2)dt between e and 1.

The first step is to differentiate the substitution.Because u=2+ln(t) we can differentiate to get du=(1/t) dt which can be rearranged to dt=t du.Once we have this we can start on the actual expression. We start by substituting both u and dt into the expression to obtain t/(t(u^2)). From this we can simplify by cancelling t from the top and bottom of the fraction to obtain 1/(u^2) which can be rewritten as simply u^-2. This is easy to integrate and it gives us (-1/u)+c. However as we are integrating with limits the "+c" is not necessary. Instead we should substitute u back in to get -1/(2+ln(t)). Then substitute the limits into the expression and subtract the lower limit from the upper limit to get (-1/(2+ln(e)))-(-1/(2+ln(1))). ln(e) is equal to 1 and ln(1) is equal to 0. Knowing this we can simplify the expression to (-1/(2+1))-(-1/(2+0)) which equals (-1/3)-(-1/2) which equals our answer which is 1/6.