Improper Integrals Question 4

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

$ -\dfrac{1}{9}$

$ \dfrac{1}{9}$

$ \dfrac{1}{3}$

Divergent

Approach

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

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