What is the integral of 1/(x log(x)) dx?

What is the integral of 1/(x log(x)) dx?

• A. log(|x|) + C
• B. log(|log(x)|) + C
• C. x log(x) + C
• D. 1/x + C

Answer: (B)

To find the integral of \( \frac{1}{x \log(x)} \, dx \), we can utilize a substitution method that simplifies the integral. Consider the substitution \( u = \log(x) \). This leads to \( du = \frac{1}{x} \, dx \), which means \( dx = x \, du = e^u \, du \). Substituting \( u \) back into the integral, we can rewrite the expression: \[ \int \frac{1}{x \log(x)} \, dx = \int \frac{1}{x u} \cdot x \, du = \int \frac{1}{u} \, du \] Now, the integral of \( \frac{1}{u} \) is a well-known integral that results in \( \log |u| + C \). Since we set \( u = \log(x) \), substituting back we obtain: \[ \log(|\log(x)|) + C \] This confirms that the correct answer is indeed the integral of \( \frac{1}{x \log(x)} \, dx \). This method clarifies why the

To find the integral of ( \frac{1}{x \log(x)} , dx ), we can utilize a substitution method that simplifies the integral.

Consider the substitution ( u = \log(x) ). This leads to ( du = \frac{1}{x} , dx ), which means ( dx = x , du = e^u , du ). Substituting ( u ) back into the integral, we can rewrite the expression:

[

\int \frac{1}{x \log(x)} , dx = \int \frac{1}{x u} \cdot x , du = \int \frac{1}{u} , du

]

Now, the integral of ( \frac{1}{u} ) is a well-known integral that results in ( \log |u| + C ). Since we set ( u = \log(x) ), substituting back we obtain:

[

\log(|\log(x)|) + C

]

This confirms that the correct answer is indeed the integral of ( \frac{1}{x \log(x)} , dx ).

This method clarifies why the