Find the integral of 1/x dx.

Find the integral of 1/x dx.

• A. ln(x) + C
• B. ln|x| + C
• C. 1/x + C
• D. x^(-1) + C

Answer: (B)

Integrating the function \( \frac{1}{x} \) results in the natural logarithm function, but it is essential to account for the domain of \( x \). The natural logarithm, \( \ln(x) \), is only defined for positive values of \( x \). However, \( \frac{1}{x} \) is defined for both positive and negative values (excluding zero). To express the integral more generally and accommodate both positive and negative values of \( x \), we use the absolute value of \( x \). The correct integral is therefore: \[ \int \frac{1}{x} \, dx = \ln|x| + C \] where \( C \) represents the constant of integration. This formulation ensures that the logarithm remains valid for all values of \( x \) except for zero, which is where the function \( \frac{1}{x} \) is undefined. The other expressions, while they either return \( \ln(x) + C \) or forms that may seem similar, do not correctly represent the entire domain of the integral of \( \frac{1}{x} \). Thus, the accepted solution is indeed \( \ln|x

Integrating the function ( \frac{1}{x} ) results in the natural logarithm function, but it is essential to account for the domain of ( x ). The natural logarithm, ( \ln(x) ), is only defined for positive values of ( x ). However, ( \frac{1}{x} ) is defined for both positive and negative values (excluding zero).

To express the integral more generally and accommodate both positive and negative values of ( x ), we use the absolute value of ( x ). The correct integral is therefore:

[

\int \frac{1}{x} , dx = \ln|x| + C

]

where ( C ) represents the constant of integration. This formulation ensures that the logarithm remains valid for all values of ( x ) except for zero, which is where the function ( \frac{1}{x} ) is undefined.

The other expressions, while they either return ( \ln(x) + C ) or forms that may seem similar, do not correctly represent the entire domain of the integral of ( \frac{1}{x} ). Thus, the accepted solution is indeed ( \ln|x