Question

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

• A. $(ae{)}^x+C$
• B. $\frac{(ae{)}^x}{log(ae)}+C$
• C. $\frac{(e{)}^x}{1+loga}+C$
• D. $Noneofthese$

Answer 1

Answer: (B)

$\int e^{xloga}{e}^{x}dx$
$I=\int a^x\cdot e^{x}dx\dots \left(i\right)$
$\Rightarrow I=\left[e^x\cdot \frac{a^x}{lo{g}_{e}a}-\int e^x\cdot \frac{a^x}{lo{g}_{e}a}dx\right]$
$\Rightarrow I=\frac{e^x\cdot a^x}{lo{g}_{e}a}-\frac{1}{lo{g}_{e}a}\int e^x\cdot a^x\cdot dx$
$\Rightarrow I=\frac{e^x\cdot a^x}{lo{g}_{e}a}-\frac{1}{lo{g}_{e}a}\int e^x\cdot a^x\cdot dx$
$\Rightarrow I=\frac{e^x\cdot a^x}{lo{g}_{e}a}-\frac{1}{lo{g}_{e}a}\cdot I$ [from Eq.t(i)]
$\Rightarrow \left(\frac{1+lo{g}_{e}a}{lo{g}_{e}a}\right)$
$I=\frac{e^x\cdot a^x}{lo{g}_{e}a}$
$\Rightarrow \left(lo{g}_{e}e+lo{g}_{e}a\right)$
$I=e^x\cdot a^x$
$\Rightarrow I=\frac{{\left(ea\right)}^x}{log\left(ae\right)}+c$