$\int a^{x}\,dx$ (with $a>0,\ a\neq 1$) equals
A$a^{x}\log a+C$
B$x\,a^{x-1}+C$
C$\dfrac{a^{x}}{\log a}+C$
D$\dfrac{a^{x+1}}{x+1}+C$
Answer & Solution
Correct answer: C. $\dfrac{a^{x}}{\log a}+C$
1. From the standard list, $\dfrac{d}{dx}\left(\dfrac{a^{x}}{\log a}\right)=a^{x}$.
2. Therefore $\int a^{x}\,dx=\dfrac{a^{x}}{\log a}+C$.
3. Option A is the derivative, not the integral.
_Source: syllabus Mathematics Ch 7 "Integrals", p.290_
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