$∫\frac{x+e^{2x}}{x^2+e^{2x}}=$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{2}log|x^2+e^{2x}|+C$
$I=∫\frac{x+e^{2x}}{x^2+e^{2x}}dx$
let $y=x^2+e^{2x}$
$dy=2x+2e^{2x}dx$
so $\frac{dy}{2}=x+e^{2x}dx$
$I = \frac{1}{2}\int\frac{dy}{y}=\frac{1}{2}\log y +C$
$=\frac{1}{2}\log(x^2+e^{2x})+C$