The block diagram of a feedback control system is shown in the figure:

 

The output of the $G_2$ path is not simply $G_{2}X(s)$ because $G_2$ does not receive $X(s)$ directly.

Look at the first summing junction:

$$X(s)⟶\boxed{{\displaystyle \mathrm{\Sigma }}}⟶E(s)$$

The signal after the summing junction is $E(s)$. Both $G_1$ and $G_2$ receive this same $E(s)$:

$$\boxed{{\displaystyle E(s)\to \{\begin{array}{l}{G}_1\\ G_2\end{array}}}$$

Let the output of the first summing junction be $E(s)$.

Because the feedback is negative:

$$E(s)=X(s)-H(s)G_1(s)E(s)$$

Hence,

$$E(s)\left[1+G_1H\right]=X(s)$$

$$E(s)=\frac{X(s)}{1+G_1H}$$

At the output summing junction, the two signals are added:

$$Y(s)=G_1E(s)+G_2E(s)$$

$$Y(s)=(G_1+G_2)E(s)$$

Therefore,

$$\boxed{\frac{Y(s)}{X(s)} =\frac{G_1+G_2}{1+G_1H}}$$

Answer: (c)