Question

$\int e^{xloga }e^{x} dx$ is equal to

• A. $(ae)^x+C$
• B. $\frac{(ae)^x}{log(ae)}+C$
• C. $\frac{(e)^x}{1+log a}+C$
• D. $None\,of\,these$

Answer 1

Answer: (B)

$\frac{(ae)^x}{log(ae)}+C$

Answer 2

$\int e^{x\,log\,a} e^{x}dx$
$I=\int a^{x}\cdot e^{x} dx \ldots\left(i\right)$
$\Rightarrow I=\left[e^{x}\cdot\frac{a^{x}}{log_{e} a}-\int e^{x}\cdot\frac{a^{x}}{log_{e} a} dx\right]$
$\Rightarrow I=\frac{e^{x}\cdot a^{x}}{log_{e}a}-\frac{1}{log_{e} a}\int e^{x}\cdot a^{x}\cdot dx$
$\Rightarrow I=\frac{e^{x}\cdot a^{x}}{log_{e} a }-\frac{1}{log_{e} a} \int e^{x}\cdot a^{x}\cdot dx$
$\Rightarrow I=\frac{e^{x}\cdot a^{x}}{log_{e} a}-\frac{1}{log_{e} a}\cdot I$ [from Et(i)]
$\Rightarrow \left(\frac{1+log_{e} a}{log_{e} a}\right)$
$I=\frac{e^{x}\cdot a^{x}}{log_{e} a}$
$\Rightarrow \left(log_{e} e+log_{e} a\right)$
$I=e^{x}\cdot a^{x}$
$\Rightarrow I=\frac{\left(ea\right)^{x}}{log\left(ae\right)}+c$