Answer
$E(p)= \frac{\lambda}{4\pi \varepsilon _0 b} \left[ \frac{L-a}{\sqrt{(L-a)^2+b^2}}+\frac{a}{\sqrt{a^2+b^2}} \right ]$
Work Step by Step
$E(p)=\int_{-a}^{L-a} \frac{\lambda b}{4\pi \varepsilon _0 (x^2+b^2)^\frac{3}{2}} dx$
Integrate without limits
$I=\int \frac{\lambda b}{4\pi \varepsilon _0 (x^2+b^2)^\frac{3}{2}} dx$
$\begin{Bmatrix}
x=b \tan\theta & x^2=b^2 \tan^2\theta \\
\frac{dx}{d\theta}=b\sec^2\theta & dx=b\sec^2\theta {d\theta}
\end{Bmatrix}$⁴
$=\int\frac{ \lambda b\sec^2\theta b }{4\pi \varepsilon _0 (b^2 \tan^2\theta+b^2)^\frac{3}{2}}{d\theta}$
$=\frac{\lambda}{4\pi \varepsilon _0} \int\frac{b^2 \sec^2\theta}{(b^2( \tan^2\theta+1))^\frac{3}{2}}{d\theta}$
$=\frac{\lambda}{4\pi \varepsilon _0} \int\frac{b^2 \sec^2\theta}{(b^2\sec^2\theta)^\frac{3}{2}}{d\theta}$
$=\frac{\lambda}{4\pi \varepsilon _0} \int\frac{b^2 \sec^2\theta}{b^3\sec^3\theta}{d\theta}$
$=\frac{\lambda}{4\pi \varepsilon _0} \int\frac{1}{b\sec\theta}{d\theta}$
$=\frac{\lambda}{4\pi \varepsilon _0 b} \int\cos \theta{d\theta}$
$=\frac{\lambda}{4\pi \varepsilon _0 b} (\sin \theta)+C$
$\tan\theta=\frac{x}{b}$
$\sin\theta =\frac{x}{\sqrt{x^2+b^2}}$
$=\frac{\lambda}{4\pi \varepsilon _0 b} \left[ \frac{x}{\sqrt{x^2+b^2}} \right ] + C$
Use limits
$E(p)=\frac{\lambda}{4\pi \varepsilon _0 b} \left[ \frac{x}{\sqrt{x^2+b^2}} \right ]_{-a}^{L-a}$
$=\frac{\lambda}{4\pi \varepsilon _0 b} \left[ \frac{L-a}{\sqrt{(L-a)^2+b^2}}-\frac{(-a)}{\sqrt{(-a)^2+b^2}} \right ]$
$=\frac{\lambda}{4\pi \varepsilon _0 b} \left[ \frac{L-a}{\sqrt{(L-a)^2+b^2}}+\frac{a}{\sqrt{a^2+b^2}} \right ]$
