Answer
$\frac{2GMm}{a^{2}}[1-\frac{x}{\sqrt {x^{2}+a^{2}}}]$
Work Step by Step
We can view the disk as being made up of smaller concentric disks of radius $r$ and thickness $dr.$
The area of each ring is $dA = 2πrdr$ and the mass is $dM = \frac{M}{πa^{2}}dA = \frac{2Mr}{a^{2}}dr$.
Then the force exerted by this ring on the mass $m$ is . . .
$dF = \frac{(GmdM)x}{(r^{2}+x^{2})^{3/2}} = \frac{2GmMx}{a^{2}}\frac{rdr}{(r^{2}+x^{2})^{3/2}} $
The total force is the contribution of all the forces from all the disks. So we integrate from 0 to a:
$F = \int dF =\int_{0}^{a} \frac{2GmMx}{a^{2}}\frac{rdr}{(r^{2}+x^{2})^{3/2}} $
Clearly, $\frac{2GmMx}{a^{2}}$ is a constant quantity, so we bring it outside the integral.
$F = \frac{2GmMx}{a^{2}}\int_{0}^{a} \frac{rdr}{(r^{2}+x^{2})^{3/2}} $
Then we evaluate the integral (using the substitution $u^{2} = r^{2} + x^{2}$). This yields . . .
$F = \frac{2GmMx}{a^{2}}[\frac{1}{x}-\frac{1}{\sqrt {a^{2}+x^{2}}}] = \frac{2GMm}{a^{2}}[1-\frac{x}{\sqrt {x^{2}+a^{2}}}] $
To verify this result and to check if it gives us the correct result as x grows very large, we can rewrite the term within brackets as . . .
$\frac{x}{\sqrt {x^{2} + a^{2}}} = \frac{1}{\sqrt {1 + (a/x)^{2}}} = (1+(a/x)^2)^{-1/2}$
Using the binomial expansion, if $x>>a$, this expression simplifies to $F=\frac{GMm}{x^{2}}$—as we should expect! So the result is correct.
