What is the result of integrating (1/x) with respect to x?

What is the result of integrating (1/x) with respect to x?

• A. ln|x| + C
• B. e^x + C
• C. x + C
• D. 1/x + C

Answer: (A)

The integration of \( \frac{1}{x} \) with respect to \( x \) leads to the natural logarithm function, specifically \( \ln|x| \). The reason for this stems from the fundamental properties of logarithmic functions and their derivatives. When differentiating \( \ln|x| \), we apply the chain rule. The derivative of \( \ln|x| \) is \( \frac{1}{x} \). This fact confirms that when you reverse the differentiation process through integration, the integral of \( \frac{1}{x} \) indeed yields \( \ln|x| + C \), where \( C \) represents the constant of integration. This function, \( \ln|x| \), is defined for all non-zero values of \( x \), which is important to consider when evaluating the integral over its domain. The absolute value notation, \( |x| \), ensures that the logarithm remains valid for both positive and negative values of \( x \) (except at zero, which is undefined). In summary, the integration of \( \frac{1}{x} \) correctly gives you \( \ln|x| + C \), cementing its role in

The integration of ( \frac{1}{x} ) with respect to ( x ) leads to the natural logarithm function, specifically ( \ln|x| ). The reason for this stems from the fundamental properties of logarithmic functions and their derivatives.

When differentiating ( \ln|x| ), we apply the chain rule. The derivative of ( \ln|x| ) is ( \frac{1}{x} ). This fact confirms that when you reverse the differentiation process through integration, the integral of ( \frac{1}{x} ) indeed yields ( \ln|x| + C ), where ( C ) represents the constant of integration.

This function, ( \ln|x| ), is defined for all non-zero values of ( x ), which is important to consider when evaluating the integral over its domain. The absolute value notation, ( |x| ), ensures that the logarithm remains valid for both positive and negative values of ( x ) (except at zero, which is undefined).

In summary, the integration of ( \frac{1}{x} ) correctly gives you ( \ln|x| + C ), cementing its role in