Integral Calculus
86361
\(\int e^{x \operatorname{loga}} e^{x} d x\) is equal to :
1 \(\frac{\mathrm{a}^{\mathrm{x}}}{\log \mathrm{ae}}+\mathrm{c}\)
2 \(\frac{\mathrm{e}^{\mathrm{x}}}{1+\log _{\mathrm{e}} \mathrm{a}}+\mathrm{c}\)
3 \((a)^{\mathrm{x}}+\mathrm{c}\)
4 \(\frac{(\mathrm{ae})^{\mathrm{x}}}{\log _{\mathrm{e}} \mathrm{ae}}+\mathrm{c}\)
5 \(\frac{a^{x} e^{x}}{\log _{x} a}+c\)
Explanation:
(D) : Given, \(I=\int \mathrm{e}^{x \log a} \mathrm{e}^{\mathrm{x}} \mathrm{dx}\)
\(=\int e^{\log a^{x}} e^{x} d x=\int a^{x} e^{x} d x=\int(a e)^{x} d x\)
\(=\frac{(a e)^{x}}{\log _{e} a e}+c \quad\left(\because \int a^{x} d x=\frac{a^{x}}{\log a}+c\right)\)