Can we Prove without wolfram that $\lim_{x\to \infty}\frac12 \log\frac{x^2-x+1}{x^2+x+1}=-3.985527186 \times 10^{-16}$

Question

While calculating an another integral, I had to calculate the following integral

$$\int_{0}^{\infty}\frac{x^2-1}{x^4+x^2+1}dx$$

Indefinite integral of above function yields $$\frac12 \log\frac{x^2-x+1}{x^2+x+1}$$

Plotting the function $\frac12 \log\frac{x^2-x+1}{x^2+x+1}$ on desmos, I can see that it is exactly $0$ at $x=0$ (As given function cuts $x$-axis at origin) but it is $\to 0$ as $x\to \infty$.

Wolfram says Integral is about $-3.985527186 \times 10^{-16}$.That means It the limit of the function $\frac12 \log\frac{x^2-x+1}{x^2+x+1}$ at $x=\infty$

My question is How can we prove without wolfram that $$\lim_{x\to \infty}\frac12 \log\frac{x^2-x+1}{x^2+x+1}=-3.985527186 \times 10^{-16}$$

Answer 1

\begin{align} \lim_{x\to \infty} \ln\frac{x^2-x+1}{x^2+x+1} &= \lim_{x\to \infty} \ln\left( \frac{x^3+1}{x^3-1} \cdot \frac{x-1}{x+1} \right) \\ &=\lim_{x\to \infty} \ln\left(\frac{x^3+1}{x^3-1}\right) +\lim_{x\to \infty}\ln\left(\frac{x-1}{x+1}\right) \\ &=\lim_{x\to \infty} \ln\left(\frac{1+\tfrac1{x^3}}{1-\tfrac1{x^3}}\right) +\lim_{x\to \infty}\ln\left(\frac{1-\tfrac1x}{1+\tfrac1x}\right) =0 . \end{align}

Or just

\begin{align} \lim_{x\to \infty} \ln\frac{x^2-x+1}{x^2+x+1} &= \lim_{x\to \infty} \ln\left( 1-\frac{2x}{x^2+x+1} \right) \\ &= \lim_{x\to \infty} \ln\left( 1-\frac{2}{x+1+\tfrac1x} \right) =0 . \end{align}

Answer 2

We could just use l'Hopital's rule to evaluate the limit: $$\begin{align} \lim_{x\to \infty} \ln\frac{x^2-x+1}{x^2+x+1} &=\ln\left(\lim_{x\to \infty}\frac{x^2-x+1}{x^2+x+1}\right)\\ &=\ln\left(\lim_{x\to \infty}\frac{2x-1}{2x+1}\right)\\ &=\ln\left(\lim_{x\to \infty}\frac{2}{2}\right)=\ln1=0.\\ \end{align}$$

Answer 3

$$I(t)=\int_{0}^{t}\frac{x^2-1}{x^4+x^2+1}dx=\frac12 \log\left(\frac{t^2-t+1}{t^2+t+1}\right)=\frac12 \log\left(\frac{1-\frac 1t+\frac 1 {t^2} }{1-\frac 1t+\frac 1 {t^2} }\right)$$ $$I(t)=-\frac{1}{t}+\frac{2}{3 t^3}+O\left(\frac{1}{t^5}\right)$$