TITLE: Proving an estimate for this integral
QUESTION [7 upvotes]: How can I show that$$\sqrt[3]6>\int_1^\infty\frac{(1+x)^{1/3}}{x^2}\mathrm dx?$$

REPLY [2 votes]: The integral can be written as
$$\int_1^{\infty} {1 \over x^{5 \over 3}} \bigg(1 + {1 \over x}\bigg)^{1 \over 3}\,dx$$
By Taylor expanding, $(1 + {1 \over x})^{1 \over 3} = 1 + {1 \over 3x} +$ error term, where the error term is negative by the Lagrange form of the remainder. So the integral is less than
$$\int_1^{\infty} {1 \over x^{5 \over 3}} + {1 \over 3 x^{8 \over 3}}\,dx$$
$$= {3 \over 2} + {1 \over 3}\cdot{3 \over 5}$$
$$= 1.7$$
$$ < \sqrt[3]6$$