Calculating the following integral $\int^{\infty}_{2}\frac{x-1}{x^3+4x+7}$ I'm trying to find the solution of the following integral, to check for convergence or divergence. Though, I'm finding the integral itself difficult to resolve. I welcome any feedback towards my approach.
$$\int^{\infty}_{2}\frac{x-1}{x^3+4x+7}$$
My approach:
$$\int^{\infty}_{2}\frac{x-1}{x^3+4x+7} \implies \frac{1}{2}\int^{\infty}_{2}\frac{2x-4}{x^3+4x+7}+\frac{1}{2}\int^{\infty}_{2}\frac{2}{x^3+4x+7}$$
Then to solve for the left-side first, I'm substiting u for $x^2-4x+7$, so the denominator looks like this: $xu$
Then to find x, such that: $x = \frac{u-7}{x-4}$
To get:
$$\frac{1}{2}\int^{\infty}_{2}\frac{du}{\frac{u-7}{x-4}\cdot u}$$
Though, I'm not sure if my apporach is valid, hence I would really enjoy some support towards my approach.

Write $x^3+4x+7=(x-a)((x-b)^2+c^2)$ with $a\approx-1.25538,\,b\approx0.627692,\,c\approx2.2764$. Then$$\begin{align}\frac{x-1}{x^3+4x+7}&\equiv\frac{A}{x-a}+\frac{B(x-b)+C}{(x-b)^2+c^2}\\\iff x-b+b-1&\equiv A((x-b)^2+c^2)+(x-b+b-a)(B(x-b)+C)\\&=(A+B)(x-b)^2+((b-a)B+C)(x-b)+Ac^2+(b-a)C\\\iff A=-B&=\frac{a-1}{(b-a)^2},\\C&=\frac{b-1}{b-a}.\end{align}$$One antiderivative is $A\ln\tfrac{|x-a|}{\sqrt{(x-b)^2+c^2}}+\tfrac{C}{c}\arctan\tfrac{x-b}{c}$, so the integral is $A\ln\tfrac{\sqrt{(2-b)^2+c^2}}{|2-a|}+\tfrac{C}{c}\arctan\tfrac{c}{2-b}$.
