calculus - Integral from $0$ to $infty$ of $frac{1}{3}lnleft(frac{x+1}{sqrt{x^2-x+1}}right)+frac{1}{sqrt 3}arctan left(frac{2x-1}{sqrt 3} right)$

Evaluate the integral $$\int_0^\infty \left( \frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)+\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right) \right) dx$$

I have read about improper integrals, and seen that I should do just like I would do with a and b, only that now, when I have $\infty$ I should take the limit. But I do not know hot to evaluate the limit of $$\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)$$ and $$\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right)$$

Answer

As $x\to\infty$, $\log\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)\to0$. However, $\arctan\left(\frac{2x-1}{\sqrt3}\right)\to\frac\pi2$. Thus, the integrand tends to $\frac\pi{2\sqrt3}$, so the integral cannot converge.

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To compute the limits:

$$\lim_{x\to\infty}\log\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)=\log\left(\lim_{x\to\infty}\frac{1+1/x}{\sqrt{1-1/x+1/x^2}}\right)$$
and substitute $u=\frac{2x-1}{\sqrt3}$ to get
$$\lim_{x\to\infty}\frac1{\sqrt3}\arctan\left(\frac{2x-1}{\sqrt3}\right)=\frac1{\sqrt3}\lim_{u\to\infty}\arctan(u)$$