(Comparison Theorem) Why is x/(x^3+1) convergent on interval 0 to infinity?

[Kaede_N9]

∫0->∞ x/(x^3 + 1) dx. Use comparison theorem to determine whether the integral is convergent of divergent.

Homework Equations

None.

The Attempt at a Solution

∫0->∞ x/(x^3 + 1) dx

= ∫0->∞ x/(x^3) dx
= ∫0->∞ 1/(x^2) dx

From my class I learned that
∫1->∞ 1/(x^2) dx , is convergent

But now that the interval begins from 0 to infinity
∫0->∞ 1/(x^2) dx is divergent!

Although my professor, and as well as the back of the book, tells me that ∫0->∞ x/(x^3+1) dx is convergent.

This must mean that I did something wrong... what would that be?

 

[micromass]

∫0->∞ x/(x^3 + 1) dx

= ∫0->∞ x/(x^3) dx

Uuuh, why?

 

[Kaede_N9]

Uuuh, why?

1/(x+1)

if x = ∞
1/(∞+1) ≈ 0
0 ≈ 1/(∞+1) ≈ 1/(∞) ≈ 0

 

[HallsofIvy]

You can compare $x/(x^3+ 1)< x/x^3= 1/x^2$ to show that the integral from 1 to infinity is finite. And now, because the original integrand is finite on the interval from 0 to 1, that entire integral is convergent.

 

[Kaede_N9]

You can compare $x/(x^3+ 1)< x/x^3= 1/x^2$ to show that the integral from 1 to infinity is finite. And now, because the original integrand is finite on the interval from 0 to 1, that entire integral is convergent.

For the original integrand, we know that it is finite on the interval from 0 to 1 by plugging in 0 and 1? It seems like the original integrand does not have an asymptote like 1/x^2 does.

 

[SammyS]

For the original integrand, we know that it is finite on the interval from 0 to 1 by plugging in 0 and 1? It seems like the original integrand does not have an asymptote like 1/x^2 does.

That is not enough to show that it's finite on the whole interval [0, 1].