Question Number 150932 by maged last updated on 16/Aug/21
$$\mathrm{I}=\int\frac{\mathrm{cosx}\:−\:\mathrm{2sinx}}{\mathrm{e}^{\mathrm{2x}} −\mathrm{sinx}}\mathrm{dx}\overset{?} {=} \\ $$
Answered by Olaf_Thorendsen last updated on 16/Aug/21
$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{cos}{x}−\mathrm{2sin}{x}}{{e}^{\mathrm{2}{x}} −\mathrm{sin}{x}}\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\left(\mathrm{2}{e}^{\mathrm{2}{x}} −\mathrm{2sin}{x}\right)−\left(\mathrm{2}{e}^{\mathrm{2}{x}} −\mathrm{cos}{x}\right)}{{e}^{\mathrm{2}{x}} −\mathrm{sin}{x}}\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\left(\mathrm{2}−\frac{\mathrm{2}{e}^{\mathrm{2}{x}} −\mathrm{cos}{x}}{{e}^{\mathrm{2}{x}} −\mathrm{sin}{x}}\right)\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{2}{x}−\mathrm{ln}\mid{e}^{\mathrm{2}{x}} −\mathrm{sin}{x}\mid+\mathrm{C} \\ $$
Commented by maged last updated on 16/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\: \\ $$