# 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

To integrate something by substitution (also known as the change-of-variable rule), we need to select a function

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

- Let
#u = ln(x)# , then#(du)/dx = 1/x# - this is a standard derivative. - 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# - 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# - 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# - Integrate normally. The answer you get for this "new" integral will be
*exactly the same answer*as the original integral.