What is the integral ∫ (1/x^2) dx?

What is the integral ∫ (1/x^2) dx?

• A. -1/x + C
• B. 1/x + C
• C. 2/x + C
• D. -2/x + C

Answer: (A)

To find the integral of \( \int \frac{1}{x^2} \, dx \), we can rewrite the integrand using the power rule. The function \( \frac{1}{x^2} \) can be expressed as \( x^{-2} \). Using the power rule of integration, which states that \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) for any real number \( n \neq -1 \), we can apply it here. Here, \( n = -2 \). Applying the power rule: \[ \int x^{-2} \, dx = \frac{x^{-2 + 1}}{-2 + 1} + C = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C \] Thus, the integral evaluates to \( -\frac{1}{x} + C \), which is why the correct answer is indeed the first option. The constant \( C \) represents the constant of integration, which arises because the operation of integration can yield multiple functions that differ by a constant. Therefore, \( -\frac

To find the integral of ( \int \frac{1}{x^2} , dx ), we can rewrite the integrand using the power rule. The function ( \frac{1}{x^2} ) can be expressed as ( x^{-2} ).

Using the power rule of integration, which states that ( \int x^n , dx = \frac{x^{n+1}}{n+1} + C ) for any real number ( n \neq -1 ), we can apply it here. Here, ( n = -2 ).

Applying the power rule:

[

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

]

Thus, the integral evaluates to ( -\frac{1}{x} + C ), which is why the correct answer is indeed the first option.

The constant ( C ) represents the constant of integration, which arises because the operation of integration can yield multiple functions that differ by a constant. Therefore, ( -\frac