Determine whether the improper integral converges or diverges.
-$$\int _ { 1 } ^ { \infty } \frac { \sqrt { 6 x + 2 } } { x ^ { 2 } }$$
A) Converges
B) Diverges

Question

Quizplus · Accepted Answer

The integral converges because as $x$ approaches infinity, the function $\frac{\sqrt{6x+2}}{x^2}$ behaves like $\frac{1}{x^{3/2}}$, which converges when integrated from 1 to infinity since the exponent of $x$ in the denominator is greater than 1.

Determine Whether the Improper Integral Converges or Diverges ∫1∞6x+2x2\int _ { 1 } ^ { \infty } \frac { \sqrt { 6 x + 2 } } { x ^ { 2 } }∫1∞​x26x+2​​

Determine Whether the Improper Integral Converges or Diverges $\int _ { 1 } ^ { \infty } \frac { \sqrt { 6 x + 2 } } { x ^ { 2 } }$