Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 18

2∫(-2excos(2x))dx+2exsin(2x)-1excos(2x)=2(-2∫1excos(2x)dx)+2exsin(2x)-1excos(2x)
Use integration by parts:
int (u)dv = uv-intvdu

Let u = cos(2x) and dv = dx/e^x
Then,
du = -2sin2xdx
v = int(e^-x)dx = -1/e^x

Thus,
int (1/e^xcos(2x))dx =(cos(2x)⋅(-1/e^x)−int((-1/e^x)⋅(-2sin(2x))dx))=(-int(2/e^xsin(2x)dx-1/e^xcos(2x)))
Use the constant multiple rule
-int(2/e^xsin(2x)dx-1/e^xcos(2x))=-(2int(1/e^xsin(2x)dx)-1/e^xcos(2x))
Use integration by parts again for the first part
Let u=sin(2x) and dv=dx/e^x
Then,
du = 2cos(2x)dx and v = int(e^-x)dx = -1/e^-x
Integral becomes

-2int(1/e^xsin(2x)dx-1/e^xcos(2x))=-2(sin(2x)(-1/e^x)-int((-1/e^x)2cos(2x)dx))
-1/e^xcos(2x)=-2(-int((-2/e^xcos(2x))dx-1/e^xsin(2x)))-1/e^xcos(2x)
Apply constant multiple rule
2int((-2/e^xcos(2x))dx)+2/e^xsin(2x)-1/e^xcos(2x)=2(-2int(1/e^xcos(2x)dx))+2/e^xsin(2x)-1/e^xcos(2x)
Simplify
int(1/e^xcos(2x))dx=1/(5e^x)(2sin(2x)-cos(2x))+C