Question

Solve:
$\int \frac{e^x}{2e^{2x}+3e^x+5}dx$

Answer 1

$\int \frac{e^x}{{2e}^{2x}+{3e}^x+5}dx=\frac{1}{2}\int \frac{e^{x}dx}{e^{2x}+\frac{3}{2}{e}^x+\frac{5}{2}}$

$=\frac{1}{2}\int \frac{e^{x}dx}{{\left(e^x\right)}^2+2.\frac{3}{9}.e^x+{\left(\frac{3}{4}\right)}^2-{\left(\frac{3}{9}\right)}^2+\frac{5}{2}}$

$=\frac{1}{2}\int \frac{e^{x}dx}{{(e^x+\frac{3}{4})}^2+\frac{31}{16}}$

Let $t=e^x+\frac{3}{4}\Rightarrow dt=e^{x}dx$

$=\frac{1}{2}\int \frac{dt}{t^2+{\left(\frac{\sqrt{31}}{4}\right)}^2}$

$=\frac{1}{2}.\frac{1}{\left(\frac{\sqrt{31}}{4}\right)}{\tan }^{-1}\left(\frac{t}{\frac{\sqrt{31}}{4}}\right)+c$

$=\frac{2}{\sqrt{31}}{\tan }^{-1}\left(\frac{4(e^x+\frac{3}{4})}{\sqrt{31}}\right)+c$

$=\frac{2}{\sqrt{31}}{\tan }^{-1}\left(\frac{4e^x+3}{\sqrt{31}}\right)+c$

$\int \frac{e^x}{2e^{2x}+3e^x+5}dx=\frac{2}{\sqrt{31}}{\tan }^{-1}\left(\frac{4e^x+3}{\sqrt{31}}\right)+c$