What does ∫ x^n dx evaluate to when n ≠ -1?

What does ∫ x^n dx evaluate to when n ≠ -1?

• A. (1/(n+1))x^(n+1) + C
• B. (1/n)x^n + C
• C. (n)x^(n-1) + C
• D. (1/(n-1))x^(n-1) + C

Answer: (A)

The integral of \( x^n \) when \( n \neq -1 \) evaluates to \( \frac{1}{n+1} x^{n+1} + C \) due to the power rule of integration. This rule states that for any real number \( n \) other than -1, the integral can be computed by increasing the exponent by one (from \( n \) to \( n+1 \)) and then dividing by the new exponent. Thus, if you start with \( x^n \), when you increase the exponent as per the rule, you get \( x^{n+1} \). Since you need to account for the factor resulting from this change in exponent, you divide by the new exponent, \( n+1 \). Hence, the expression is written as \( \frac{1}{n+1} x^{n+1} \), and you add a constant \( C \) to represent the constant of integration. This method of finding the integral applies broadly in calculus and aligns with the basic principles governing the integration of power functions.

The integral of ( x^n ) when ( n \neq -1 ) evaluates to ( \frac{1}{n+1} x^{n+1} + C ) due to the power rule of integration. This rule states that for any real number ( n ) other than -1, the integral can be computed by increasing the exponent by one (from ( n ) to ( n+1 )) and then dividing by the new exponent.

Thus, if you start with ( x^n ), when you increase the exponent as per the rule, you get ( x^{n+1} ). Since you need to account for the factor resulting from this change in exponent, you divide by the new exponent, ( n+1 ). Hence, the expression is written as ( \frac{1}{n+1} x^{n+1} ), and you add a constant ( C ) to represent the constant of integration.

This method of finding the integral applies broadly in calculus and aligns with the basic principles governing the integration of power functions.