Question

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

Answer 1

$\begin{array}{c}\int \frac{e^x}{e^{2x}-3e^x+1}\\ Let{e}^x=t\Rightarrow e^x\frac{dt}{dx}\\ \int \frac{dt}{t^2-3t+1}=\int \frac{dt}{{\left(t-\frac{3}{2}\right)}^2-\frac{5}{4}}\\ =\int \frac{dt}{{\left(t-\frac{3}{2}\right)}^2-{\left(\frac{\sqrt{5}}{2}\right)}^2}\\ =\frac{1}{\sqrt{5}}\ln \left(\frac{t-\frac{3}{2}-\frac{\sqrt{5}}{2}}{t-\frac{3}{2}+\frac{\sqrt{5}}{2}}\right)+C\\ =\frac{1}{\sqrt{5}}\ln \left(\frac{2e^x-3-\sqrt{5}}{2e^x-3+\sqrt{5}}\right)+C\end{array}$

Hence,

solved.