If $∫xe^x\,cosx\,dx=f(x)+c$, then f(x) is equal to

$\frac{e^x}{2}\{(x-1)sinx+x\,cosx\}$

Here I = Real part of $∫xe^{(1+i)x}dx$ and

$∫xe^{(1+i)x}dx=\frac{xe^{(1+i)x}}{(1+i)}-\frac{e^{(1+i)x}}{(1+i)}dx=\frac{xe^{(1+i)x}}{(1+i)}-\frac{e^{(1+i)x}}{(1+i)^2}$

$=e^{(1+i)x}\begin{Bmatrix}\frac{x(1+i)-1}{(1+i)^2}\end{Bmatrix}$

= e^x [cos x + i sin x] $[\frac{(x-1)ix}{1+2i-1}]=\frac{e^x}{-2}\{cosx-sinx\}\{(x-1)+ix\}$

Hence $I =\frac{e^x}{-2}$ [(1 - x) sin x – x cos x] =$\frac{e^x}{2}$ [x cos x + (x – 1) sin x] .

Hence (B) is the correct answer.