\(\int\displaystyle (e^{x\log a}+e^{a\log x})\, dx\) is equal to (where \(a \gt 1\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{a^x}{\log a}+\frac{x^{a+1}}{a+1}+C\): C is an arbitrary constant
Since \(e^{\log t} = t\), both exponentials simplify:
\(e^{x\log a} = e^{\log a^x} = a^x\)
\(e^{a\log x} = e^{\log x^a} = x^a\)
So
\( \displaystyle \int\left(e^{x\log a} + e^{a\log x}\right)dx = \int a^x\,dx + \int x^a\,dx \)
Using \(\int a^x\,dx = \frac{a^x}{\log a}\) (valid as \(a \gt 1\), so \(\log a \ne 0\)) and the power rule:
\( \displaystyle = \frac{a^x}{\log a} + \frac{x^{a+1}}{a+1} + C \)
Why the other options are incorrect
Option (2) \(\displaystyle(\log a) a^x+\frac{x^{a+1}}{a+1}+C\): C is an arbitrary constant — Incorrect. \((\log a)\,a^x\) is the derivative of \(a^x\), not its integral.
Option (3) \(\displaystyle\frac{a^x}{a+1}+\frac{x^{a+1}}{\log a}+C\): C is an arbitrary constant — Incorrect. The two denominators are interchanged; the derivative of \(\frac{a^x}{a+1}\) is \(\frac{(\log a)\,a^x}{a+1}\), not \(a^x\).
Option (4) \((a+1)a^x+(\log a)x^{a+1}+C\): C is an arbitrary constant — Incorrect. Its derivative is \((a+1)(\log a)\,a^x + (a+1)(\log a)\,x^a\), not \(a^x + x^a\).