This is a derivation of the deflection of light by a point mass, closely following the derivation in Weinberg's book, Gravitation and Cosmology (1972).

$$d\tau^2=Bdt^2-Adr^2-r^2(d\theta^2+\sin^2\theta d\phi^2),$$

with

$$B = A^{-1}=\left(1-\frac{2GM}{r}\right).$$

A test particle characterized by 4-velocity $u^\mu$ is following a geodesic trajectory given by the equation of motion $du^\mu/d\tau+\Gamma^\mu_{\alpha\beta}u^\alpha u^\beta=0$. In the Schwarzschild metric, this equation translates into

\begin{align*}
\dfrac{d^2t}{d\tau^2}+\dfrac{B'}{B}\dfrac{dt}{d\tau}\dfrac{dr}{d\tau}&=0,\\
\dfrac{d^2r}{d\tau^2}+\dfrac{B'}{2A}\left(\dfrac{dt}{d\tau}\right)^2+\dfrac{A'}{2A}\left(\dfrac{dr}{d\tau}\right)^2-\dfrac{r}{A}\left(\dfrac{d\phi}{d\tau}\right)^2&=0,\\
\dfrac{d^2\phi}{d\tau^2}+\dfrac{2}{r}\dfrac{dr}{d\tau}\dfrac{d\phi}{d\tau}&=0,
\end{align*}

where the prime denotes differentiation with respect to $r$. The first of these equations can be rearranged and integrated:

\begin{align*}
B\dfrac{d^2t}{d\tau^2}+\dfrac{dB}{d\tau}\dfrac{dt}{d\tau}&=0,\\
\dfrac{dt}{d\tau}&=\dfrac{C}{B},
\end{align*}

where $C$ is some constant. Since at infinity, $B=dt/d\tau=1$, we must have $C=1$.

Meanwhile, the third equation of motion can be integrated directly:

$$r^2\dfrac{d\phi}{d\tau}=J,$$

where $J$, another integration constant, can be identified as the angular momentum per unit mass. Using these results in the second equation of motion, we get

$$\dfrac{d^2r}{d\tau^2}+\dfrac{A'}{2A}\left(\dfrac{dr}{d\tau}\right)^2-\dfrac{J^2}{Ar^3}+\dfrac{B'}{2A}\left(\dfrac{1}{B}\right)^2=0.$$

Multiplication by $2Adr/d\tau$ yields

$$2A\dfrac{dr}{d\tau}\dfrac{d^2r}{d\tau^2}+A'\left(\dfrac{dr}{d\tau}\right)^3-2\dfrac{dr}{d\tau}\dfrac{J^2}{r^3}+\dfrac{dr}{d\tau}\dfrac{B'}{B^2}=0,$$

or

$$\dfrac{d}{d\tau}\left[A\left(\dfrac{dr}{d\tau}\right)^2+\dfrac{J^2}{r^2}-\dfrac{1}{B}\right]=0,$$

or

$$A\left(\dfrac{dr}{d\tau}\right)^2+\dfrac{J^2}{r^2}-\dfrac{1}{B}=-{\cal E},$$

where ${\cal E}$ is another integration constant. In the limit of large $r$ ($A=B=1$, $dr/dt=dr/d\tau=v$) we get

$${\cal E}=1-v^2.$$

Using our earlier result for $r^2d\phi/d\tau$ in the preceding equation, we get

$$A\left(\dfrac{dr}{d\phi}\dfrac{J}{r^2}\right)^2+\dfrac{J^2}{r^2}-\dfrac{1}{B}=-{\cal E},$$

or

$$\dfrac{A}{r^4}\left(\dfrac{dr}{d\phi}\right)^2+\dfrac{1}{r^2}=-\dfrac{\cal E}{J^2}+\dfrac{1}{J^2B}.$$

At closest approach, $r=r_0$ and $dr/d\phi=0$. Thus

$$J=r_0\sqrt{\dfrac{1}{B_0}-{\cal E}},$$

where $B_0=1-2GM/r_0$ is the value of $B$ at $r=r_0$. We can solve the preceding equation for $\phi$:

$$\phi=\pm\int\dfrac{\sqrt{A}}{r^2\left[\dfrac{1}{r_0^2}\left(\dfrac{1}{B}-{\cal E}\right)\left(\dfrac{1}{B_0}-{\cal E}\right)^{-1}-\dfrac{1}{r^2}\right]^{1/2}}~dr.$$

For an ultrarelativistic particle, ${\cal E}=1-v^2\sim 0$ and can be omitted, leaving us with

$$\phi=\pm\int\dfrac{\sqrt{A}}{r\left[\dfrac{r^2}{r_0^2}\dfrac{B_0}{B}-1\right]^{1/2}}~dr.$$

The square rooted expression in the denominator can be simplified to first order:

\begin{align*}
\dfrac{r^2}{r_0^2}\dfrac{B_0}{B}-1
&=\dfrac{r^2}{r_0^2}\left[1+2GM\left(\dfrac{1}{r}-\dfrac{1}{r_0}\right)\right]-1\\
&=\left(\dfrac{r^2}{r_0^2}-1\right)\left[1-2GM\left(\dfrac{r}{r_0(r+r_0)}\right)\right]
\end{align*}

So to first order, we can write the integral as

$$\phi=\pm\int\dfrac{r_0}{r\sqrt{r^2-r_0^2}}\left[1+\dfrac{GM}{r}+GM\dfrac{r}{r_0(r+r_0)}\right]~dr.$$

This is an elementary integral that can be readily evaluated. Now for a particle coming from infinity and going to infinity, the deflection angle will be given by

$$\Delta\phi=2\left|\int_{r_0}^\infty\dfrac{r_0}{r\sqrt{r^2-r_0^2}}\left[1+\dfrac{GM}{r}+GM\dfrac{r}{r_0(r+r_0)}\right]~dr\right|-\pi=\frac{4GM}{r_0}.$$

After restoring the speed of light, we get therefore

$$\Delta\phi=\dfrac{4GM}{c^2r_0}.$$

Substituting $G=6.674\times 10^{-11}~{\rm m}^3/{\rm s}^2{\rm kg}$, $M=2\times 10^{30}~{\rm kg}$, $r_0=6.95\times 10^8~{\rm m}$ (and of course $c=3\times 10^8~{\rm m}/{\rm s}$), parameters that characterize a ray of light grazing the Sun, we get

$$\Delta\phi\sim 1.76".$$

It is also interesting to contemplate where the factor of 4 is coming from. In the Schwarzschild metric, both time and space have curvature, characterized by the metric coefficients $B$ and $A$. Newtonian gravity amounts to ignoring the spatial curvature; i.e., setting $A=1$. And indeed, if we set the $\sqrt{A}$ term in the integral expression for $\Delta\phi$ to 1, we get exactly half the predicted deflection, $2GM/c^2r_0$. Which demonstrates explicitly that it is because of the presence of spatial curvature, which has no nonrelativistic analog, that the bending of light has the predicted magnitude. Therefore, observations like Eddington's (concerns about the data quality notwithstanding) are indeed genuine tests of general relativity.