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

Answer 1

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:

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Answer 2

Ezekiel Alexander

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 $2\sqrt{4} - 2\sqrt{1} = 4 - 2 = 2$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.