\( \int x ^ { n } d x = \frac { x ^ { n + 1 } } { n + 1 } + c \)

You've correctly provided the formula for the indefinite integral of \( x^n \) with respect to \( x \). Here's a detailed explanation: \[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \] **Where:** - \( n \) is any real number except \( n = -1 \). - \( C \) is the constant of integration, representing an arbitrary constant that arises because the integral is indefinite. **Important Consideration:** - **Case when \( n = -1 \):** The formula above does **not** apply when \( n = -1 \) because it would involve division by zero. Instead, when \( n = -1 \), the integral of \( x^{-1} \) (which is \( \frac{1}{x} \)) is: \[ \int \frac{1}{x} \, dx = \ln|x| + C \] **Summary:** - For \( n \neq -1 \): \[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \] - For \( n = -1 \): \[ \int \frac{1}{x} \, dx = \ln|x| + C \] **Example:** Let's compute \( \int x^{2} \, dx \): \[ \int x^{2} \, dx = \frac{x^{3}}{3} + C \] **Another Example with \( n = -2 \):** \[ \int x^{-2} \, dx = \int \frac{1}{x^{2}} \, dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C \] Feel free to ask if you have more questions about integration or any other mathematical concepts!