Question

How can I decide whether this integral is convergent or divergent?

∫(0,1) 1/(x^3-2x^2+x)

Answer 1

I have a phrase they say if: The integral
∫ (0.1) 1/x^p
converges if p <1 and diverges if p ≥ 1

I'm almost sure it's the one I should use, but I do not know how.

Answer 2

I would try to "factorize",
\[\int_0^1\frac{1}{x^3-2x^2+x}dx=\int_0^1{\left(\frac{1}{x}-\frac{1}{x-1}+\frac{1}{(x-1)^2}\right)dx}\]
And apply the theorem term by term,

Answer 3

Thank you. If each term diverges so does the integral it too? You know what to do if some of the terms converge and other diverge?

Answer 4

If any one term diverges, the whole integral does.

Answer 5

"∫ (0.1) 1/x^p
converges if p <1 and diverges if p ≥ 1"

Isn't this the other way around? It diverges if p =< 1 and converges if p>1.

Answer 6

I dont think so...

Wacth the attachment.

Answer 7

Agent0smith Perhaps you are thinking of the integral: \[\int\limits_{1}^{∞}dx/x^p\] converges if p > 1 and diverges if p ≤ 1.
???

Answer 8

Yeah, it appears you're right. http://www.sosmath.com/calculus/improper/convdiv/convdiv.html

Answer 9

If one of the terms diverges, then the integral diverges. In fact, you can see that the last term is the one that make the true divergence, numerically spaking. An alternative way to test the divergence is calculating the integral. Then you will see that each term diverges.

If one term converges but one diverges, then the integral diverges.

Answer 10

John have won solve this integral:

\[\int\limits_{0}^{1}1/(x^3+x^2-2x)\]?

Answer 11

maybe someone else can help me?

Answer 12

\[\Large \int\limits_0^1{\left(\frac{1}{x}-\frac{1}{x-1}+\frac{1}{(x-1)^2}\right)dx}\]

All terms of this are divergent.

Answer 13

Thank you Smith

Answer 14

I do not think the equation is correct. It's the same as previously!?!

Answer 15

The indefinite integral can be calculated,
\[\ln|x|-\ln|x-1|-\frac{1}{x-1}+K\]
You can check the limits x->0, and x->1, you will find the expression is divergent.

Answer 16

@RFJ-86 the two are equal, you can check it by getting the left side over a common denominator x(x-1)^2 (John_ES used partial fractions to break it up)

\[\LARGE {\frac{1}{x}-\frac{1}{x-1}+\frac{1}{(x-1)^2}} = \frac{1}{x^3-2x^2+x}\]

Answer 17

\[\Large \int\limits\limits_0^1{\left(\frac{1}{x}-\frac{1}{x-1}+\frac{1}{(x-1)^2}\right)dx} \]
From this all you need to realize is that 1/x is divergent, so the whole integral is (if one term diverges to infinity, does it matter if the other terms converge?)

Answer 18

@agent0smith

I agree with this equation
\[\frac{ 1 }{ x }-\frac{ 1 }{ x-1 }+\frac{ 1 }{ (x-1)^2 }=\frac{ 1 }{ x^3-2x^2+x }\]
But this is one i not true:
\[\frac{ 1 }{ x }-\frac{ 1 }{ x-1 }+\frac{ 1 }{ (x-1)^2 }\ \neq \frac{ 1 }{ x^3+x^2-2x }\]
I need a way to "factorize" this equation
\[\frac{ 1 }{ x^3+x^2-2x }\]

Answer 19

But your original integral was the 1/(x^3-2x^2+x)...

For the one you just posted, pull out an x
\[\Large \frac{ 1 }{ x(x^2+x-2) }\] then factor the brackets \[\Large \frac{ 1 }{ x(x+2)(x-1) }\]
Now you should be able to make partial fractions.

Answer 20

Thank you @agent0smith