Question

$\int \frac{dx}{1+3e^x+2e^{2x}}.$

• A. $\log \frac{e^x(1-e^x)}{{(1+2e^x)}^2}.$
• B. $\log \frac{e^x(1+e^x)}{{(1+e^x)}^2}.$
• C. $\log \frac{e^x(1+e^x)}{{(1+2e^x)}^2}.$
• D. $\log \frac{e^x(1+e^x)}{{(1-2e^x)}^2}.$

Answer 1

The correct option is C $\log \frac{e^x(1+e^x)}{{(1+2e^x)}^2}.$

Let $I=\int \frac{dx}{1+3e^x+2e^{2x}}$

Substitute $e^x=t\Rightarrow e^{x}dx=dt$

$I=\int \frac{1}{t(1+3t+2t^2)}dt=\int (\frac{1}{t}+\frac{1}{1+t}-\frac{2}{1+2t})dt$
$=\log t+\log (1+t)-2\log (1+2t)=\log e^x+\log (1+e^x)-2\log (1+2e^x)$

$=\log \frac{e^x(1+e^x)}{{(1+2e^x)}^2}$