From the given
#f(x)=int (2-xe^x)(e^x+cos x) dx#
My solution begins with the expansion of the integrand by multiplying the two binomials
#f(x)=int (2-xe^x)(e^x+cos x) dx#
#f(x)=int (2e^x+2cos x-xe^(2x)-xe^xcos x) dx#
#f(x)=#
#int 2e^x* dx+ int 2cos x* dx-int xe^(2x)* dx- int xe^xcos x* dx#
Integrate each term of the integrand separately
First term
#int 2e^x* dx=color(red)(2e^x+C_0)#
Second term
#int 2cos x* dx=color(red)(2sin x+C_1)#
Third term-by integration by parts #int u* dv=uv-int v*du#
Let #u=x# and #dv=e^(2x)dx# and #v=1/2e^(2x)# and #du=dx#
#int xe^(2x)* dx=color(red)(1/2xe^(2x)-1/4e^(2x)+C_2)#
Fourth term-by integration by parts
Let #u=x# and #dv=e^x cos x* dx# and #v=1/2e^xsin x+1/2e^xcos x# and #du=dx#
#int xe^xcos x* dx=color(red)(1/2xe^xsin x+1/2xe^xcos x-1/2e^xsin x+C_3)#
Add all these integration
#f(x)=2e^x+2sin x-(1/2xe^(2x)-1/4e^(2x))-(1/2xe^xsin x+1/2xe^xcos x-1/2e^xsin x)+C#
#color(red)(f(x)=2e^x+2sin x-1/2xe^(2x)+1/4e^(2x)-1/2xe^xsin x-1/2xe^xcos x+1/2e^xsin x+C)#
But #f(0)=2#
#f(0)=2=2e^0+2sin (0)-1/2(0)e^(2(0))+1/4e^(2(0))-1/2(0)e^(0)sin (0)-1/2(0)e^(0)cos (0)+1/2e^(0)sin (0)+C#
#2=2(1)+2(0)-0+1/4*(1)-0-0+0+C#
#color(red)(C=-1/4)#
Final answer
#color(blue)(f(x)=2e^x+2sin x-1/2xe^(2x)+1/4e^(2x)-1/2xe^x sin x-1/2xe^x cos x+1/2e^xsin x-1/4)#
God bless...I hope the explanation is useful.
