Comparison Test for Improper Integrals — Question 10

Question 10 – Solution

Step 1: Bound the denominator. For $x\ge1$, we have $x\le x^2$, so $$3x\le3x^2.$$ Adding $x^2$ gives $$x^2\le x^2+3x\le4x^2.$$ Step 2: Take reciprocals. All terms are positive, so $$\frac1{4x^2}\le\frac1{x^2+3x}\le\frac1{x^2}.$$ Step 3: Test the upper comparison. $$\int_1^\infty\frac{dx}{x^2}=1<\infty.$$ Step 4: Apply direct comparison. Since the original integrand is nonnegative and bounded above by an integrable function, the integral converges. $$\boxed{\text{converges}}$$