A uniform rod of mass M and length L is considered. Find the gravitational field intensity at point P on perpendicular bisector 

$\text{dm}=\frac{\text{M}}{\text{L}}dx$

$\text{dg}=\frac{\text{Gdm}}{\sqrt{{\text{d}}^{\text{2}}+x^2}}$

Net gravitational field at point p   ${\text{g}}_{\text{p}}={\displaystyle \int \text{dg}\cos \theta }$

${\text{g}}_{\text{p}}={\displaystyle \int \frac{\text{GMdx}}{\text{L}\sqrt{{\text{d}}^{\text{2}}+x^2}}}.\frac{\text{d}}{\sqrt{{\text{d}}^{\text{2}}+x^2}}$

${\text{g}}_{\text{p}}=\frac{\text{GMd}}{\text{L}}{\displaystyle \underset{-\text{L/2}}{\overset{+\text{L/2}}{\int }}\frac{\text{dx}}{{\left({\text{d}}^{\text{2}}+x^2\right)}^{\text{3/2}}}}$

$\tan \theta =\frac{x}{\text{d}}$

$x=\text{d}\tan \theta$

$dx=\text{d}{\sec }^2\theta d\theta$

${\text{d}}^2+x^2={\left({\text{d}}^2{\tan }^2\theta +{\text{d}}^2\right)}^{\text{3/2}}$

$={\text{d}}^3{\sec }^3\theta$

${\text{g}}_{\text{p}}=\frac{\text{GMd}}{\text{L}}{\displaystyle \underset{-\text{L/2}}{\overset{+\text{L/2}}{\int }}\frac{\text{d}{\sec }^2\theta d\theta }{{\text{d}}^3{\sec }^3\theta }}$

${\text{g}}_{\text{p}}=\frac{\text{GM}}{\text{Ld}}{\displaystyle \underset{-\text{L/2}}{\overset{+\text{L/2}}{\int }}\cos \theta d\theta }$

${\text{g}}_{\text{p}}=\frac{\text{GM}}{\text{Ld}}{\left[\sin \theta \right]}_{-\text{L/2}}^{+\text{L/2}}$

${\text{g}}_{\text{p}}=\frac{\text{GM}}{\text{Ld}}{\left[\frac{x^2}{\sqrt{{\text{d}}^{\text{2}}+x^2}}\right]}_{-\text{L/2}}^{+\text{L/2}}$

$=\frac{\text{GM}}{\text{Ld}}\left[\frac{\text{L/2}}{\sqrt{{\text{d}}^{\text{2}}+{\text{L}}^{\text{2}}\text{/4}}}\times 2\right]=\frac{4\text{GM}}{\text{L}\sqrt{4{\text{d}}^{\text{2}}+{\text{L}}^2}}$