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