$\displaystyle \int\limits_1^\infty \frac{x^2}{(x^3+2)^2}$ is

Summary Submit

Use $u$-substitution where $u=x^3+2$ to obtain:

$$\int\limits_1^\infty \frac{x^2}{(x^3+2)^2} = \lim_{b\to\infty} -\frac{1}{3(x^3+2)}\Big]_1^b$$ $$= \lim_{b\to\infty} -\frac{1}{3} \left( \frac{1}{b^3+2} - \frac{1}{1^3+2} \right)$$ $$= \lim_{b\to\infty}-\frac{1}{3}\cdot -\frac{1}{3}$$ $$= \boxed{\frac{1}{9}}$$