I assumed the square(side a) sheet to be made up of wires.$$dE=Kdq/r^2$$ The field due to a wire is : Reference $$\frac{K\lambda}{d}\left[\frac{x}{\sqrt{d^2+x^2}}\right]^{(a/2)}_{(-a/2)}=\frac{K\lambda}{d}\left[\frac{2a}{\sqrt{4d^2+a^2}}\right]$$ where $d$ is the distance between the centre of rod/wire and the point and $\lambda$ is charge density
Now for finding the field due to the square plate, I used this integral: $$\int{\left( \cfrac{2K\sigma ah}{(h^2+y^2)\sqrt{4(h^2+y^2)+a^2}}\right) dy}$$ $h$ is the height of the point above centre of plate $\sigma$ is surface charge density. And since horizontal components cancel out, $\sin\theta=h/d$ has been multiplied.
Assuming $\tan\theta=y/h$, the integral gives: $$\int{\left( \cfrac{2K\sigma a}{h\sqrt{4h^2\sec^6\theta+a^2\sec^4\theta}}\right) d\tan\theta}$$ $$=\int{\left( \cfrac{2K\sigma a}{h\sqrt{4h^2\sec^2\theta+a^2}}\right) d\theta}$$
How do I proceed further?
Are there any alternatives to evaluate that integral?
PS: Here are two images for visualization but the variables are I used are different so bear with that.
This may not be the shortest solution, but here we go. We want to compute $$ \int\frac{2K\sigma ah}{(h^2+y^2)\sqrt{4(h^2+y^2)+a^2}}dy. $$ Rearranging, this is the same as $$ 2K\sigma ah\int\frac{1}{(h^2+y^2)\sqrt{4y^2+4h^2+a^2}}dy. $$ We make the substitution $y=\frac{\sqrt{4h^2+a^2}}{2}\tan(\theta)$ with $dy=\frac{\sqrt{4h^2+a^2}}{2}\sec^2(\theta)d\theta$. Our integral becomes $$ 2K\sigma ah\frac{\sqrt{4h^2+a^2}}{2}\int\frac{\sec^2(\theta)}{(h^2+\frac{4h^2+a^2}{4}\tan^2(\theta))\sqrt{4h^2+a^2}\sec(\theta)}d\theta, $$ or $$ K\sigma ah\int\frac{\sec(\theta)}{h^2+\frac{4h^2+a^2}{4}\tan^2(\theta)}d\theta. $$ This is the same as $$ 4K\sigma ah\int\frac{\sec(\theta)}{4h^2+(4h^2+a^2)\tan^2(\theta)}d\theta. $$ Now $\sec(\theta)=\frac{1}{\cos(\theta)}$, so this integral is the same as $$ 4K\sigma ah\int\frac{1}{4h^2\cos(\theta)+(4h^2+a^2)\cos(\theta)\tan^2(\theta)}d\theta. $$ Since $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$, this is the same as $$ 4K\sigma ah\int\frac{1}{4h^2\cos(\theta)+(4h^2+a^2)\sin(\theta)\tan(\theta)}d\theta, $$ or $$ 4K\sigma ah\int\frac{\cos(\theta)}{4h^2\cos^2(\theta)+(4h^2+a^2)\sin^2(\theta)}d\theta. $$ Distributing, this yields $$ 4K\sigma ah\int\frac{\cos(\theta)}{4h^2\cos^2(\theta)+4h^2\sin^2(\theta)+a^2\sin^2(\theta)}d\theta, $$ and a trig identity gives us $$ 4K\sigma ah\int\frac{\cos(\theta)}{4h^2+a^2\sin^2(\theta)}d\theta. $$ Let $u=\sin(\theta)$, so that $du=\cos(\theta)d\theta$. With this substitution, our integral becomes $$ 4K\sigma ah\int\frac{1}{4h^2+a^2u^2}du. $$ A standard trig substitution should solve this integral. I think the solution is $$ 4K\sigma ah\frac{1}{2ha}\arctan\left(\frac{au}{2h}\right)+C=2K\sigma\arctan\left(\frac{au}{2h}\right)+C. $$ Now we have to undo our substitutions. We get first that the integral is equal to $$ 2K\sigma\arctan\left(\frac{a\sin(\theta)}{2h}\right)+C. $$ You can continue substituting in the rest for $\theta$, as I don't really want to work through the computations to simplify this.