How do you prove the convergence of $\int_1^\infty \frac{x-2}{x^3+1}\,dx$ using the comparison test?

Q: How do you prove the convergence of $\int_1^\infty \frac{x-2}{x^3+1}\,dx$ using the comparison test?

To prove convergence, split the integral into $[1,2]$ and $[2,\infty)$. On $[1,2]$, the integrand is continuous, so the finite piece is well-defined. On $[2,\infty)$, verify $0\le\frac{x-2}{x^3+1}\le\frac1{x^2}$. Since $\int_2^\infty \frac1{x^2}\,dx$ converges (as $p=2>1$), the comparison test proves convergence on the tail. Combining both pieces proves convergence from 1.

To prove convergence, split the integral into

[1,2]

and

[2,\infty)

. On

[1,2]

, the integrand is continuous, so the finite piece is well-defined. On

[2,\infty)

, verify

0\le\frac{x-2}{x^3+1}\le\frac1{x^2}

. Since

\int_2^\infty \frac1{x^2}\,dx

converges (as

p=2>1

), the comparison test proves convergence on the tail. Combining both pieces proves convergence from 1.

Conditions

The integral is $\int_1^\infty \frac{x-2}{x^3+1}\,dx$ .
The comparison function is $\frac{1}{x^2}$ .
The inequality holds for $x \ge 2$ .

Reasoning, step by step

Split the domain into $[1,2]$ and $[2,\infty)$ .
Verify that the integrand is continuous on $[1,2]$ , ensuring the finite integral exists.
For $x \ge 2$ , establish the inequality $0\le\frac{x-2}{x^3+1}\le\frac1{x^2}$ .
Confirm that $\int_2^\infty \frac1{x^2}\,dx$ converges because $p=2 > 1$ .
Apply the comparison test to conclude convergence on $[2,\infty)$ .
Combine the results to prove convergence on $[1,\infty)$ .

Example

For $x\ge2$ , $0\le\frac{x-2}{x^3+1}\le\frac1{x^2}$ . The comparison proves convergence on $[2,\infty)$ . On $[1,2]$ the integrand is continuous, so that finite piece is well-defined; combining both pieces proves convergence from 1.

Common misconceptions

Ignoring the finite interval $[1,2]$ and assuming the comparison holds everywhere.
Failing to check that the integrand is nonnegative on the tail before applying the test.
Assuming the integral converges just because the denominator grows faster than the numerator, without formal comparison.

Watch the explanation

YouTubeComparison Test for Improper Integrals

5:16 – 7:14Watch this moment ↗

Connected concepts

Improper integrals →

How do you prove the convergence of $\int_1^\infty \frac{x-2}{x^3+1}\,dx$ using the comparison test?

Conditions

Reasoning, step by step

Example

Common misconceptions

Watch the explanation

Connected concepts

Explore next

Related questions

How do you prove the convergence of ∫1∞x−2x3+1 dx\int_1^\infty \frac{x-2}{x^3+1}\,dx∫1∞​x3+1x−2​dx using the comparison test?

Conditions

Reasoning, step by step

Example

Common misconceptions

Watch the explanation

Connected concepts

Explore next

Related questions

What does the comparison test not determine about an improper integral?+

When does the p-integral ∫1∞x−p dx\int_1^\infty x^{-p}\,dx∫1∞​x−pdx converge or diverge?+

How does the comparison test decide convergence or divergence for improper integrals with nonnegative functions?+

How do you prove the convergence of $\int_1^\infty \frac{x-2}{x^3+1}\,dx$ using the comparison test?