Question

Evaluate:$\int \frac{e^{2x}}{e^{4x}-e^{2x}+2}$ dx

Answer 1

Consider the given function:

$I=\int \frac{e^{2x}}{e^{4x}-e^{2x}+2}dx$

$let{e}^{2x}=t$

$2e^{2x}dx=dt$

$thenI=\int \frac{e^{2x}}{e^{4x}-e^{2x}+2}dx$

$=\frac{1}{2}\int \frac{1}{t^2-t+2}dt$

$=\frac{1}{2}\int \frac{1}{t^2-t+{\left(\frac{1}{2}\right)}^2-{\left(\frac{1}{2}\right)}^2+2}dt$

$=\frac{1}{2}\int \frac{1}{{(t-\frac{1}{2})}^2+\frac{7}{4}}dt$

$=\frac{1}{2}\int \frac{1}{{(t-\frac{1}{2})}^2+{\left(\frac{\sqrt{7}}{2}\right)}^2}dt$

$=\frac{1}{2}.\frac{1}{\frac{\sqrt{7}}{2}}{\tan }^{-1}\left(\frac{t-\frac{1}{2}}{\frac{\sqrt{7}}{2}}\right)$

$=\frac{1}{\sqrt{7}}{\tan }^{-1}\left(\frac{2t-1}{\sqrt{7}}\right)$

$=\frac{1}{\sqrt{7}}{\tan }^{-1}\left(\frac{2e^{2x}-1}{\sqrt{7}}\right)$

Hence this is the answer.