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

Sep 6, 2014

Since dx has been placed twice, I'm going to assume that the equation should read ${\int}_{e}^{{e}^{4}} \frac{\mathrm{dx}}{x \cdot \sqrt{\ln \left(x\right)}}$. In this case, a good way to find the integral is by substitution, letting $u = \ln \left(x\right)$.

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 $\tan \left(x\right)$ we can say that $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$ and select $u = \cos \left(x\right)$. The derivative of $u$ is also within the original equation.)

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

1. Let $u = \ln \left(x\right)$, then $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$ - this is a standard derivative.
2. Substitute these two new functions into the equation:
${\int}_{e}^{{e}^{4}} \frac{\mathrm{dx}}{x \cdot \sqrt{\ln \left(x\right)}} = {\int}_{e}^{{e}^{4}} \frac{1}{x \cdot u} \mathrm{dx} = {\int}_{e}^{{e}^{4}} \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$
3. Find new terminals - this is a crucial step, because we're changing the variable!
$\text{When "x = e, u = ln(e) = 1 " and when } x = {e}^{4} , u = \ln \left({e}^{4}\right) = 4$
4. Substitute these new terminals in, and "cancel out" the two $\mathrm{dx}$ terms:
${\int}_{e}^{{e}^{4}} \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx} = {\int}_{1}^{4} \frac{\mathrm{du}}{u} = {\int}_{1}^{4} {u}^{- 1} \mathrm{du}$
5. Integrate normally. The answer you get for this "new" integral will be exactly the same answer as the original integral.