I have used the substitution $u= x+1  $ then $du =dx$ to evaluate the following integral in closed form and since $\arctan$ is connected to $x^2+1$ as it is a derivative of it. $$\int_{1}^{\infty} \frac{\arctan (x+1)}{x^2+4} dx$$  But i didn't come up to it's closed form the  me the closed form for it , and all my attempts gives me this approach : $$\int_{1}^{\infty} \frac{\arctan (x+1)}{x^2+4} dx=\frac{\pi^2}{8}-\frac{3\pi}{8}\arctan \left(\frac{1}{2}\right)$$ But this is not the same result with , Any way ?
I was playing around with some integrals and noticed that some integrals of the form:  $$I(a,b,c)=\int_a^\infty \frac{\arctan(x+b)}{x^2+c}dx$$ Does have a closed form. I am trying to find for what constant $c$ will this work. In case you wonder why only $c$ is problematic, I will try to show by an example.  $$I=I(1,3,16)=\int_1^\infty \frac{\arctan(x+3)}{x^2+16}dx$$ Let's start by letting $x-1=t\,$ thus: $$I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})}}{t^2+2t+17}dt$$ With $\displaystyle{t=\frac{17}{y}\rightarrow dt=-\frac{17}{y^2}dy}$ $$I=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{y}+4}\right)}}{\left(\frac{17}{y}\right)^2 +\frac{34}{y}+17 }\frac{17}{y^2}dy\overset{y=t}=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}{t^2+2t+17}dt$$ $$2I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}}{t^2+2t+17}dt$$ $${\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}=\arctan\left(\frac{\color{blue}{t+4}+\color{red}{\frac{17}{t}+4}}{1-(\color{blue}{t+4})\left(\color{red}{\frac{17}{t}+4}\right)}\right)$$ $$=\arctan\left(\frac{x^2+8x+17}{x}\frac{x}{-4(x^2+8x+17}\right)=\pi-\arctan\left(\frac14\right)$$ Above follows since the original integral is positive so we take $\arctan(-x)$ as $\pi-\arctan x $ and therefore getting a negative answer will not be an issue. $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\int_0^\infty \frac{1}{t^2+2t+17}dt$$ Well, now the inner integral is not hard to compute and the final answer happens to be: $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\frac14\arctan\left(\frac{t+1}{4}\right)\bigg|_0^\infty =\frac{\pi^2}{16}-\frac{3\pi}{16}\arctan\left(\frac14\right)+\frac18\arctan^2\left(\frac14\right) $$ There are more examples that I found by checking and try such as:  $$I(1,2,9)=\int_1^\infty \frac{\arctan(x+2)}{x^2+9}dx$$ $$I(2,1,6)=\int_2^\infty \frac{\arctan(x+1)}{x^2+6}dx$$ $$I(2,2,13)=\int_2^\infty \frac{\arctan(x+2)}{x^2+13}dx$$ And so on... All those can be solved by the same method: First substitute $x-a=t$, then let $t=\frac{\alpha}{y}$, where $\alpha$ is the "free of x" coefficient from the denominator.   The problem is that I tried more than $100$ combinations to get those integrals which is not that nice. How can we "smartly" find $c$ so that $I(a,b,c)$ is evaluable by symmetry? Or put in other word what should be $c$ if one wants to compute by symmetry $I(7,13,c)$?
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