Question

Solve :

$\int e^{x\log a}\cdot e^{x}dx$

Note : In the options all logarithmic functions are having base as $e$

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

Answer 1

The correct option is C $\frac{e^{x(\log a+1)}}{(loga+1)}+c$

Given $\int e^{xloga}.e^x.dx$

$=\int e^{x(loga+1)}dx$

$=\frac{e^{x(loga+1)}}{(loga+1)}+c$