Science:Math Exam Resources/Courses/MATH101/April 2011/Question 04 (c)/Solution 1

First notice that attempting to evaluate this integral is long and tedious. Instead we attempt to find an integral which we know converges or diverges, and compare our integral to that integral. Observe that on the given interval, $[1,\infty )$ we have that $x^{2}+x>x^{2}$ and so ${\frac {\sqrt {x}}{x^{2}+x}}<{\frac {\sqrt {x}}{x^{2}}}$. Hence,

${\begin{aligned}\int _{1}^{\infty }{\frac {\sqrt {x}}{x^{2}+x}}dx&<\int _{1}^{\infty }{\frac {\sqrt {x}}{x^{2}}}dx\\&=\int _{1}^{\infty }{\frac {1}{x^{3/2}}}dx\\\end{aligned}}$

and since we know that the integral $\int _{1}^{\infty }{\frac {1}{x^{p}}}dx$ converges whenever $p>1$, we know that our integral does indeed converge since $3/2>1$.