Compute ∫ a^x dx for a > 0, a ≠ 1.

Compute ∫ a^x dx for a > 0, a ≠ 1.

• A. ln(a) · a^x
• B. a^x
• C. x a^x
• D. a^x / ln(a)

Answer: (D)

The key idea is that exponentials of the form a^x have a derivative a^x ln a. To reverse that derivative when integrating, you divide by the constant ln a. A quick way to see it is to use substitution: let u = a^x, so du = a^x ln a dx. Then ∫ a^x dx = ∫ u · dx = ∫ u · (du/(u ln a)) = (1/ln a) ∫ du = a^x/ln a + C. Since a>0 and a ≠ 1, ln a is a nonzero constant, so the antiderivative is a^x/ln a + C.

The key idea is that exponentials of the form a^x have a derivative a^x ln a. To reverse that derivative when integrating, you divide by the constant ln a. A quick way to see it is to use substitution: let u = a^x, so du = a^x ln a dx. Then ∫ a^x dx = ∫ u · dx = ∫ u · (du/(u ln a)) = (1/ln a) ∫ du = a^x/ln a + C. Since a>0 and a ≠ 1, ln a is a nonzero constant, so the antiderivative is a^x/ln a + C.