$\int e^x\left(\frac{x^4+4x^3+64}{{(x+4)}^2}\right)dx=e^{x}f\left(x\right)+C,$Given f (0) = 6 then the value of  $7f\left(3\right)$  is _______(C is constant of integration)

$$\begin{gathered} \int e^x\left(\frac{x^4+4x^3+64}{{(x+4)}^2}\right)dx=\int e^x[\frac{x^3}{x+4}+\frac{64}{{(x+4)}^2}]dx \\ =\int e^x(\frac{x^3+64-64}{x+4}+\frac{64}{{(x+4)}^2})dx=\int e^x(x^2-4x+16-\frac{64}{x+4}+\frac{64}{{(x+4)}^2})dx \\ =e^x[(x^2-4x+16-(2x-4)+2)-\frac{64}{x+4}]+C \\ f\left(x\right)=x^2-6x+22-\frac{64}{x+4} \\ f\left(0\right)=22-16=6 \\ f\left(3\right)=13-\frac{64}{7}=\frac{27}{7} \\ 7f\left(3\right)=27 \end{gathered}$$