How do you find the integral #int_e^(e^4)dx/(x*sqrt(ln(x)))dx# ?

1 Answer
Sep 6, 2014

Since dx has been placed twice, I'm going to assume that the equation should read #int_e^(e^4)dx/(x*sqrt(ln(x)))#. In this case, a good way to find the integral is by substitution, letting #u = ln(x)#.

To integrate something by substitution (also known as the change-of-variable rule), we need to select a function #u# so that its derivative also forms part of the original equation. (For example, when we try to antidifferentiate #tan(x)# we can say that #tan(x) = sin(x)/cos(x)# and select #u = cos(x)#. The derivative of #u# is also within the original equation.)

To find the integral of your function, we do the following:

  1. Let #u = ln(x)#, then #(du)/dx = 1/x# - this is a standard derivative.
  2. Substitute these two new functions into the equation:
    #int_e^(e^4)dx/(x*sqrt(ln(x))) = int_e^(e^4)1/(x*u)dx = int_e^(e^4)1/u*(du)/dxdx#
  3. Find new terminals - this is a crucial step, because we're changing the variable!
    #"When "x = e, u = ln(e) = 1 " and when "x=e^4, u = ln(e^4)=4#
  4. Substitute these new terminals in, and "cancel out" the two #dx# terms:
    #int_e^(e^4)1/u*(du)/dxdx = int_1^4(du)/u = int_1^4u^(-1)du#
  5. Integrate normally. The answer you get for this "new" integral will be exactly the same answer as the original integral.