I have a big doubt about direction of electric field. When I try to compute a Electric Field over a point at (0,2) from the linear charge in the $x$ axis from (-5,0) to (0,0) with a density linear charge $\sigma=+20 $nC/m`, the Electric Field is $\vec{E}=-55.6i+83.55j$. This $-55.6i$ don't let me fall sleep. Question : Why the "minus" sign?, must the vector $\vec{E}$ be $\vec{E}=+55.6i+83.55j$ because the linear charge is positive?,
$dq = \sigma*ds$
$\overset{\to }{u}=\frac{\left(x\overset{\to }{i}+2\overset{\to }{j}\right)}{\sqrt{x^2+2^2}}$
$\text{dE}=\frac{\text{dq}}{4 r^2 \pi \epsilon _o}\overset{\to }{u}$
$\frac{\sigma }{4 \pi \epsilon _o}\int _{-5}^0\frac{1}{\left(x^2+2^2\right)^{3/2}}dx*\left(x\overset{\to }{i}+2\overset{\to }{j}\right)=-55.6i+83.55j$
Your calculation has a sign error.
If you consider the $x$ component of the electric force on the unit positive charge, it will be given by:
$\displaystyle dF_x = k\frac{e\sigma dx}{4+x^2} \cos \theta = k\frac{e\sigma dx}{4+x^2} \frac {|x|}{\sqrt{4+x^2}} = k\frac{e\sigma dx}{4+x^2} \frac {-x}{\sqrt{4+x^2}} \\= - ke\sigma x(4+x^2)^{-\frac 32}dx$
because, for that range of $x$ values, the absolute value $|x|$ is given by $-x$. Once you do this, you should get the correct answer.
(By the way, $\theta$ is the angle, measured counterclockwise of the electric force vector on the unit charge from the horizontal axis, $e$ is the magnitude of a unit charge (you divide this away for the electric field strength), $\sigma$ is the charge density, a constant, and $k$ is the usual constant of proportionality involving $\pi$ and the permittivity $\varepsilon_0$).