Convergence of $\int_1^\infty \frac{dt}{t^4+t^2+1}$ I have to calculate $\displaystyle\int_1^\infty \frac{dt}{t^4+t^2+1}$ and i have as a hint:
Divide by $t^2+t+1$
Then I get that $(t^2+t+1)(t^2-t+1)= t^4+t^2+1$
I could do this by partial fractions but it would be tedious...
Is there any other way that I am not seeing?
 A: HINT:
Notice, $$\int_{1}^{\infty}\frac{dt}{t^4+t^2+1}=\lim_{z\to \infty}\int_{1}^{z}\frac{dt}{t^4+t^2+1}$$
$$=\lim_{z\to \infty}\int_{1}^{z}\frac{\left(\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+1}$$
$$=\frac{1}{2}\lim_{z\to \infty}\int_{1}^{z}\frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}+1}\ dt$$
$$=\frac{1}{2}\lim_{z\to \infty}\left(\int_{1}^{z}\frac{\left(1+\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+1}\ dt-\int_{1}^{z}\frac{\left(1-\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+1}\ dt\right)$$
$$=\frac{1}{2}\lim_{z\to \infty}\left(\int_{1}^{z}\frac{\left(1+\frac{1}{t^2}\right)dt}{\left(t-\frac{1}{t}\right)^2+3}\ dt-\int_{1}^{z}\frac{\left(1-\frac{1}{t^2}\right)dt}{\left(t+\frac{1}{t}\right)^2-1}\ dt\right)$$
$$=\frac{1}{2}\lim_{z\to \infty}\left(\int_{1}^{z}\frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+(\sqrt{3})^2}\ dt-\int_{1}^{z}\frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^2-(1)^2}\ \right)$$
I hope you can take it from here 
A: Hints:
Do decompose the fraction using the given factorization. You will get two fractions with a first degree numerator and second degree denominator.
By adding a suitable constant to the numerator, you will turn it to the derivative of the denominator, making the fractions easy.
Then you need to compensate with two integrands of the form
$$\frac 1{t^2\pm t+1}.$$
By a simple shift of the variable, you transform this into
$$\frac1{t^2+c},$$
which gives an $\arctan$ or $\text{artanh}$ depending on the sign of $c$.
