Electric potential due to a small element dx of rod
\(dp = \frac 1{4\pi\epsilon_0} \frac{q\,dx}{2l\sqrt{d^2 + x^2}}\)
total \(pE = \int\limits_{-1}^1 \frac1{4\pi\epsilon_0} \frac{q\,dx}{2l\sqrt{d^2 + x^2}}\)
taking \(cos\theta = \frac d{\sqrt{d^2 + x^2}}\)
\(dx=\frac{d}{sec^2\theta }d\theta\)
and \(x=- 1 \) ⇒ \(\theta = \theta _1 = tan^{-1} (\frac{-1}d)\)
and \(x= 1 \) ⇒ \(\theta = \theta _2 = tan^{-1} (\frac{1}d)\)
\(pE = \int\limits_{\theta_1}^{\theta_2} \frac 1{4\pi\epsilon_0} \frac q{2l} \frac{cos\theta }{d} \frac{d}{cos^2\theta} d\theta\)
\(= \int\limits_{\theta_1}^{\theta_2} \frac 1{4\pi\epsilon_0} \frac q{2l} \,sec\theta \,d\theta \)
\(= \frac1{4\pi\epsilon_0} \frac q{2l} ln\left(sec\theta + tan\theta \right)_{\theta_1}^{\theta_2}\)
\(=\frac1{4\pi\epsilon_0} \frac q{2l} ln\left(\frac{\sqrt{d^2 + x^2}+1}{\sqrt{d^2 + x^2}-1}\right)\)