A parameter integral equal to arctan

(a) Differentiability. Let $g(t)=\dfrac{t}{1+t+t^2+t^3}$, which is continuous on $]0,+\infty[$ (the denominator is positive for $t>0$). By the fundamental theorem of calculus, $G(x)=\int_1^x g(t)\,dt$ is $C^1$ on $]0,+\infty[$ with $G'(x)=g(x)$. Now
$$F(x)=G(x)-G(1/x).$$
The map $x\mapsto 1/x$ is $C^1$ on $]0,+\infty[$, so by composition $G(1/x)$ is $C^1$ and
$$\frac{d}{dx}G(1/x)=g(1/x)\cdot\left(-\frac{1}{x^2}\right).$$
Thus
$$F'(x)=g(x)+\frac{1}{x^2}\,g(1/x).$$

Computation. We have
$$g(x)=\frac{x}{1+x+x^2+x^3},\qquad \frac{1}{x^2}\,g(1/x)=\frac{1}{x^2}\cdot\frac{1/x}{1+1/x+1/x^2+1/x^3}.$$
Multiply numerator and denominator of $g(1/x)$ by $x^3$:
$$g(1/x)=\frac{x^2}{x^3+x^2+x+1}.$$
Hence
$$\frac{1}{x^2}\,g(1/x)=\frac{1}{x^3+x^2+x+1}.$$
Adding:
$$F'(x)=\frac{x}{1+x+x^2+x^3}+\frac{1}{1+x+x^2+x^3}=\frac{x+1}{1+x+x^2+x^3}.$$
Now factor the denominator:
$$1+x+x^2+x^3=(1+x)(1+x^2).$$
Indeed $(1+x)(1+x^2)=1+x+x^2+x^3$. Therefore
$$F'(x)=\frac{x+1}{(1+x)(1+x^2)}=\frac{1}{1+x^2}.$$

Conclusion (a). $F$ is $C^1$ on $]0,+\infty[$ and $F'(x)=\dfrac{1}{1+x^2}$.

(b) Identification. The function $x\mapsto\arctan(x)$ has derivative $\dfrac{1}{1+x^2}$ on $]0,+\infty[$. Therefore $F-\arctan$ is constant on $]0,+\infty[$:
$$F(x)-\arctan(x)=C.$$
Evaluate at $x=1$: the lower and upper bounds coincide, so $F(1)=0$. Also $\arctan(1)=\pi/4$. Hence
$$C=0-\frac{\pi}{4}=-\frac{\pi}{4},$$
i.e.
$$F(x)=\arctan(x)-\frac{\pi}{4}.$$

Special value. For $x=\sqrt3$, $\arctan(\sqrt3)=\pi/3$, so
$$F(\sqrt3)=\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}.$$

Conclusion (b). For every $x>0$,
$$\int_{1/x}^{x}\frac{t}{1+t+t^2+t^3}\,dt=\arctan(x)-\frac{\pi}{4},$$
and in particular
$$\int_{1/\sqrt3}^{\sqrt3}\frac{t}{1+t+t^2+t^3}\,dt=\frac{\pi}{12}.$$