\(\int\displaystyle\frac{2^x+3^x}{6^x}\,dx\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle -\frac{3^{-x}}{\log 3}-\frac{2^{-x}}{\log 2}+C\)
Dividing each term of the numerator by \(6^x = 2^x\cdot 3^x\):
\( \displaystyle \frac{2^x+3^x}{6^x} = \frac{1}{3^x}+\frac{1}{2^x} = 3^{-x}+2^{-x} \)
Using \(\int a^{-x}\,dx = -\frac{a^{-x}}{\log a}+C\) (from the standard result \(\int a^x\,dx = \frac{a^x}{\log a}\) and the minus sign of \(-x\)):
\( \displaystyle \int\left(3^{-x}+2^{-x}\right)dx = -\frac{3^{-x}}{\log 3}-\frac{2^{-x}}{\log 2}+C \)
Why the other options are incorrect
Option (1) \(\frac{3^{-x}}{\log 3}+\frac{2^{-x}}{\log 2}+C\) — Incorrect. It drops the minus sign that comes from the power \(-x\); its derivative is \(-3^{-x}-2^{-x}\).
Option (2) \(-3^{-x}\log 3-2^{-x}\log 2+C\) — Incorrect. This is the derivative of \(3^{-x}+2^{-x}\), not its integral; the log must divide.
Option (4) \(-\frac{2^{-x}}{\log 3}-\frac{3^{-x}}{\log 2}+C\) — Incorrect. The bases are paired with the wrong logs; \(3^{-x}\) goes with \(\log 3\).