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

Question

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

Charles Anctil · Accepted Answer

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:

Ezekiel Alexander · Answer

undefined To find the integral $ \int_{e}^{e^4} \frac{dx}{x \sqrt{\ln(x)}} $, you can use the substitution method. Let $ u = \ln(x) $, then $ du = \frac{1}{x} dx $. Substituting this into the integral, we get:

$$ \int \frac{dx}{x \sqrt{\ln(x)}} = \int \frac{1}{\sqrt{u}} du $$

Now, the integral becomes:

$$ \int \frac{1}{\sqrt{u}} du $$

This is a standard integral which can be easily evaluated. Its antiderivative is $ 2\sqrt{u} $. Substituting back $ u = \ln(x) $, we get:

$$ 2\sqrt{\ln(x)} + C $$

So, the integral $ \int_{e}^{e^4} \frac{dx}{x \sqrt{\ln(x)}} $ evaluates to $ 2\sqrt{\ln(e^4)} - 2\sqrt{\ln(e)} $ which simplifies to $ 2\sqrt{4} - 2\sqrt{1} = 4 - 2 = 2 $.