$\dfrac{1+cosx}{1+sinx}.e^{x}dx$= $\dfrac{1}{1+sinx}.e^{x}dx$+ $\dfrac{cosx}{1+sinx}.e^{x}dx$Tính nguyên hàm sau :
$\int {\dfrac{{1 + \cos x}}{{1 + \sin x}}e^x dx}$
= $\dfrac{1}{1+sinx}.e^{x}dx$+ $e^{x}.d\dfrac{1}{1+sinx}$
= $\dfrac{1}{1+sinx}.e^{x}dx$+ $\dfrac{e^{x}}{1+sinx}$ - $\dfrac{1}{1+sinx}.e^{x}dx$
= $\dfrac{e^{x}}{1+sinx}$+C