For \(a \gt 0,\ a \ne 1\), \(\int\displaystyle\frac{a^{\sqrt{x}}}{\sqrt{x}}\,dx\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2a^{\sqrt{x}}\log_a e+C\)
Let \(t = \sqrt{x}\). Then
\( \displaystyle dt = \frac{dx}{2\sqrt{x}},\quad \frac{dx}{\sqrt{x}} = 2\,dt \)
\( \displaystyle \int\frac{a^{\sqrt{x}}}{\sqrt{x}}\,dx = 2\int a^t\,dt = \frac{2a^t}{\log_e a}+C \)
By the change of base rule, \(\frac{1}{\log_e a} = \log_a e\):
\( = 2a^{\sqrt{x}}\log_a e+C \)
Why the other options are incorrect
Option (1) \(2a^{\sqrt{x}}\log_e a+C\) — Incorrect. It multiplies by \(\log_e a\) instead of dividing by it.
Option (2) \(a^{\sqrt{x}}\log_a e+C\) — Incorrect. It misses the factor 2 from \(\frac{dx}{\sqrt{x}} = 2\,dt\).
Option (4) \(\frac{a^{\sqrt{x}}}{2\log_e a}+C\) — Incorrect. The 2 is put in the denominator; it should multiply.