\( \int\displaystyle e^{x\log_{e}a}\,.\,e^{x}\,dx \) is:
[where C is an arbitrary constant]
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{a^{x}e^{x}}{1+\log_{e}a}+C \)
Since \( e^{\log_e a}=a \), the first factor is \( e^{x\log_e a}=\left(e^{\log_e a}\right)^x=a^x \). So the integrand is a single exponential:
\( \displaystyle e^{x\log_e a}\cdot e^x=a^x e^x \)
\( \displaystyle =(ae)^x \)
Using \( \int k^x\,dx=\frac{k^x}{\log_e k}+C \) with k = ae:
\( \displaystyle \int (ae)^x\,dx=\frac{(ae)^x}{\log_e(ae)}+C \)
By the product rule of logarithms, \( \log_e(ae)=\log_e a+\log_e e=1+\log_e a \). Hence
\( \displaystyle \int e^{x\log_e a}\cdot e^x\,dx=\frac{a^x e^x}{1+\log_e a}+C \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{e^{x}}{1+\log_{e}(ae)}+C \) — Incorrect. The numerator must be \( (ae)^x \), not \( e^x \), and \( 1+\log_e(ae)=2+\log_e a \) is not the right denominator.
Option (2) \( \displaystyle\frac{a^{x}}{\log_{e}(ae)}+C \) — Incorrect. The factor \( e^x \) is missing from the numerator.
Option (3) \( (ae)^{x}+C \) — Incorrect. The derivative of \( (ae)^x \) is \( (ae)^x\log_e(ae) \), not \( (ae)^x \); the division by \( \log_e(ae) \) is missing.