\(\displaystyle\int e^{3\log x}\,dx,\ x \gt 0\), is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{x^4}{4} + C\)
Since \(3\log x = \log x^3\) and \(e^{\log t} = t\), the integrand simplifies:
\(e^{3\log x} = e^{\log x^3} = x^3\)
So
\( \displaystyle \int e^{3\log x}\,dx = \int x^3\,dx = \frac{x^4}{4} + C \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{x^3}{3} + C\) — Incorrect. Its derivative is \(x^2\), not \(x^3\); the power has not been raised by 1.
Option (3) \(\displaystyle\frac{x^4}{3} + C\) — Incorrect. Its derivative is \(\frac{4x^3}{3}\), not \(x^3\); the power has been raised but the division is by 3 instead of 4.
Option (4) \(\displaystyle\frac{3x^4}{4} + C\) — Incorrect. Its derivative is \(3x^3\); this integrates \(3x^3\), as if \(e^{3\log x}\) were \(3x^3\) instead of \(x^3\).