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

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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!
The integral of $$ x^n $$ with respect to $$ x $$ is $$ \frac{x^{n+1}}{n+1} + C $$, where $$ C $$ is the constant of integration.

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

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