What is the antiderivative of 1/x?

What is the antiderivative of 1/x?

• A. ln|x| + C
• B. e^x + C
• C. ln(x) + x + C
• D. x^2 + C

Answer: (A)

The antiderivative of \( \frac{1}{x} \) is derived from the properties of logarithmic functions. Specifically, the integral \[ \int \frac{1}{x} \, dx \] results in the natural logarithm of the absolute value of \( x \), denoted as \( \ln|x| \), plus a constant of integration \( C \). The absolute value is necessary because the logarithm function is only defined for positive values, and the integration needs to handle both positive and negative values of \( x \) appropriately. This result can be confirmed by differentiating \( \ln|x| + C \). The derivative of \( \ln|x| \) yields \( \frac{1}{x} \) for \( x \neq 0 \), thus showing that our integration is valid. The other answer choices do not represent antiderivatives of \( \frac{1}{x} \). For example, the exponential function \( e^x + C \) and the polynomial \( x^2 + C \) pertain to entirely different forms of functions and their antiderivatives. Similarly, the expression \( \ln(x) + x + C \)

The antiderivative of ( \frac{1}{x} ) is derived from the properties of logarithmic functions. Specifically, the integral

[

\int \frac{1}{x} , dx

]

results in the natural logarithm of the absolute value of ( x ), denoted as ( \ln|x| ), plus a constant of integration ( C ). The absolute value is necessary because the logarithm function is only defined for positive values, and the integration needs to handle both positive and negative values of ( x ) appropriately.

This result can be confirmed by differentiating ( \ln|x| + C ). The derivative of ( \ln|x| ) yields ( \frac{1}{x} ) for ( x \neq 0 ), thus showing that our integration is valid.

The other answer choices do not represent antiderivatives of ( \frac{1}{x} ). For example, the exponential function ( e^x + C ) and the polynomial ( x^2 + C ) pertain to entirely different forms of functions and their antiderivatives. Similarly, the expression ( \ln(x) + x + C )