Q. $\int_{}^{}{e^{x\log a}e^{x}}\ dx$ is equal to
A) $\frac{a^{x}}{\log{ae}} + c$
B) $\frac{e^{x}}{1 + \log_{e}a} + c$
C) $(ae)^{x} + c$
D) $\frac{(ae)^{x}}{\log_{e}{ae}} + c$
Explanation
(d)
Let $I = \int_{}^{}{e^{x\log a}.e^{x}}\ dx = \int_{}^{}a^{x}.e^{x}\ dx$
$= \int_{}^{}(ae)^{x}dx = \frac{(ae)^{x}}{\log_{e}{ae}} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.