Answer
The work required to increase the separation of the particles from $~~x = x_1~~$ to $~~x = x_1+d~~$ is $~~\frac{G~m_1~m_2~d}{(x_1)(x_1+d)}$
Work Step by Step
Since $F(x)$ is attractive, note that $F(x) = -\frac{G~m_1~m_2}{x^2}$
We can find the potential energy function:
$U(x) = -\int~F(x)~dx$
$U(x) = -\int~(-\frac{G~m_1~m_2}{x^2})~dx$
$U(x) = \int~\frac{G~m_1~m_2}{x^2}~dx$
$U(x) = -\frac{G~m_1~m_2}{x}$
We can find the potential energy when $x = x_1$:
$U_1 = -\frac{G~m_1~m_2}{x_1}$
We can find the potential energy when $x = x_1+d$:
$U_2 = -\frac{G~m_1~m_2}{x_1+d}$
We can find $\Delta U$:
$\Delta U = U_2-U_1$
$\Delta U = (-\frac{G~m_1~m_2}{x_1+d})-(-\frac{G~m_1~m_2}{x_1})$
$\Delta U = \frac{G~m_1~m_2}{x_1}-\frac{G~m_1~m_2}{x_1+d}$
$\Delta U = \frac{(G~m_1~m_2)(x_1+d)}{(x_1)(x_1+d)}-\frac{(G~m_1~m_2)(x_1)}{(x_1)(x_1+d)}$
$\Delta U = \frac{G~m_1~m_2~d}{(x_1)(x_1+d)}$
The work required to increase the separation of the particles from $~~x = x_1~~$ to $~~x = x_1+d~~$ is $~~\frac{G~m_1~m_2~d}{(x_1)(x_1+d)}$
