$\lim_{R \to \infty} \int_0^R \frac{dx}{(x^2+x+2)^3}$ 
$$\lim_{R \to \infty} \int_0^R \frac{dx}{(x^2+x+2)^3}$$

Please help me in this integral, I've tried some substitutions, but nothing work.
Thanks in advance!
 A: First complete the square:
$$
x^2 + x + 2 = \left( x^2 + x + \tfrac{1}{4} \right) + \tfrac{7}{4} = \left( x + \tfrac{1}{2} \right)^2 + \left( \tfrac{\sqrt{7}}{2} \right)^2.
$$
Now, make the (inverse) trigonometric substitution:
$$
\tan t = \frac{x + \tfrac{1}{2}}{\tfrac{\sqrt{7}}{2}}.
$$
This choice of ratio is motivated by the sum of square expression above.  As a consequence, we have
$$
x^2 + x + 2 = \left( \tfrac{\sqrt{7}}{2} \tan t \right)^2 + \left( \tfrac{\sqrt{7}}{2} \right)^2 = \tfrac{7}{4} \left( \tan^2 t + 1 \right) = \tfrac{7}{4} \sec^2 t.
$$
Now, the differential is
$$
dx = \tfrac{\sqrt{7}}{2} \sec^2 t \, dt
$$
and the limits of integration become
$$
\begin{align}
x &= 0 &\Longleftrightarrow \quad t(0) &= \arctan \tfrac{1}{\sqrt{7}} \\
x &= R &\Longleftrightarrow \quad t(R) &= \arctan \tfrac{2R + 1}{\sqrt{7}}.
\end{align}
$$
Now, substitute:
$$
\int_0^R \frac{dx}{(x^2 + x + 2)^3} = \int_{t(0)}^{t(R)} \frac{\tfrac{\sqrt{7}}{2} \sec^2 t \, dt}{\left( \tfrac{7}{4} \sec^2 t \right)^3} = \left( \tfrac{2}{\sqrt{7}} \right)^5 \int_{t(0)}^{t(R)} \cos^4 t \, dt.
$$
Can you finish it from here, using power reducing identities?
