\( \int e^{x\log_e 7}e^x\,dx \) is equal to :
[where C is an arbitrary constant]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{7^xe^x}{1+\log_e 7}+C \)
Since \( e^{\log_e a}=a \), the first factor of the integrand is a power of 7:
\( \displaystyle e^{x\log_e 7}=e^{\log_e 7^x} \)
\( \displaystyle =7^x \)
So the integrand is \( 7^xe^x=(7e)^x \), an exponential with base 7e. With \( \int a^x\,dx=\frac{a^x}{\log_e a}+C \):
\( \displaystyle \int (7e)^x\,dx=\frac{(7e)^x}{\log_e (7e)}+C \)
\( \displaystyle \log_e(7e)=\log_e 7+\log_e e \)
\( \displaystyle =1+\log_e 7 \)
\( \displaystyle \int e^{x\log_e 7}e^x\,dx=\frac{7^xe^x}{1+\log_e 7}+C \)
Shortcut: Check by differentiating Option (2): the numerator \( (7e)^x \) gives \( (7e)^x\log_e(7e)=7^xe^x(1+\log_e 7) \), and dividing by \( 1+\log_e 7 \) leaves exactly the integrand \( 7^xe^x \).