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What is the integral of #ln(x)/x#?

Answer 1

Lets start by breaking down the function.

#(ln(x))/x = 1/x ln(x)#

So we have the two functions;

#f(x) = 1/x# #g(x) = ln(x)#

But the derivative of #ln(x)# is #1/x#, so #f(x) = g'(x)#. This means we can use substitution to solve the original equation.

Let #u = ln(x)#.

#(du)/(dx) = 1/x#

#du = 1/x dx#

Now we can make some substitutions to the original integral.

#int ln(x) (1/x dx) = int u du = 1/2 u^2 + C#

Re-substituting for #u# gives us;

#1/2 ln(x)^2 +C#

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Answer 2

Axel Alvarez

The integral of ( \frac{\ln(x)}{x} ) is a special function known as the logarithmic integral function, denoted as ( \text{Li}(x) ). It cannot be expressed in terms of elementary functions like polynomials, exponentials, or trigonometric functions. Therefore, its integral does not have a simple closed-form expression. However, it can be represented using the logarithmic integral function as follows:

[ \int \frac{\ln(x)}{x} , dx = \text{Li}(x) + C ]

Where ( \text{Li}(x) ) is the logarithmic integral function, and ( C ) is the constant of integration.

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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.