What is the result of the integral ∫ (1/x^2) dx?

What is the result of the integral ∫ (1/x^2) dx?

• A. -1/x + C
• B. ln|x| + C
• C. x^(-1) + C
• D. (1/2)x^(-2) + C

Answer: (A)

The integral of \( \frac{1}{x^2} \) can be solved using the power rule of integration. First, we rewrite \( \frac{1}{x^2} \) as \( x^{-2} \). Now, applying the power rule for integration, which states that: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{for } n \neq -1, \] we can substitute \( n = -2 \) into this formula. Therefore, we have: 1. Increment the exponent: \(-2 + 1 = -1\). 2. Divide by the new exponent: \[ \frac{x^{-1}}{-1} = -\frac{1}{x}. \] Thus, the result of the integral becomes: \[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} + C. \] This confirms that the correct answer is the expression \(-\frac{1}{x} + C\). The other options, while they might represent different functions, do not yield the correct outcome

The integral of ( \frac{1}{x^2} ) can be solved using the power rule of integration. First, we rewrite ( \frac{1}{x^2} ) as ( x^{-2} ).

Now, applying the power rule for integration, which states that:

[

\int x^n , dx = \frac{x^{n+1}}{n+1} + C \quad \text{for } n \neq -1,

]

we can substitute ( n = -2 ) into this formula. Therefore, we have:

1. Increment the exponent: (-2 + 1 = -1).

2. Divide by the new exponent:

[

\frac{x^{-1}}{-1} = -\frac{1}{x}.

]

Thus, the result of the integral becomes:

[

\int \frac{1}{x^2} , dx = -\frac{1}{x} + C.

]

This confirms that the correct answer is the expression (-\frac{1}{x} + C).

The other options, while they might represent different functions, do not yield the correct outcome