Easy integral!!

**matty888** · May 19th 2008, 04:36 AM

Solve the following integral

**mr fantastic** · May 19th 2008, 04:37 AM

Originally Posted by matty888

Solve the following integral

Use a partial fraction decomposition.

**Moo** · May 19th 2008, 04:54 AM

Strange thing because it's not defined in 1...

**mr fantastic** · May 19th 2008, 05:02 AM

Originally Posted by Moo

Strange thing because it's not defined in 1...

Good spot. You get +oo when treating it as an improper integral and taking the limit .....

**Moo** · May 19th 2008, 05:04 AM

Originally Posted by mr fantastic

Good spot. You get +oo when treating it as an improper integral and taking the limit .....

Yep... The integral doesn't exist because the integrand isn't continuous ~
I wonder if it's a typo or whatever..

**matty888** · May 19th 2008, 05:13 AM

Thanks guys,I was asked if it converges and wasnt sure about what i was getting(0),so from ur reckoning then it will diverge,would I be right in saying that??

**mr fantastic** · May 19th 2008, 05:42 AM

Originally Posted by matty888

Thanks guys,I was asked if it converges and wasnt sure about what i was getting(0),so from ur reckoning then it will diverge,would I be right in saying that??

Yes.

**Mathstud28** · May 19th 2008, 12:07 PM

Originally Posted by matty888

Solve the following integral

$\int_0^{1}\frac{dx}{1-x^2}=arctanh(x)\bigg|_0^{1}\to\infty$ [/tex]

Alternate integration method

**Mathstud28** · May 19th 2008, 12:12 PM

Originally Posted by matty888

Solve the following integral

Also $\forall{x}\in(0,1],\frac{1}{1-x^2}>\frac{1}{x^2}$

So then $\int_0^{1}\frac{1}{x^2}dx\leq\int_0^{1}\frac{1}{1-x^2}dx$

$\int_0^{1}\frac{dx}{x^2}$ is a divergent special case p-series test

So from that we see that $\int_0^{1}\frac{dx}{1-x^2}$ diverges

**Moo** · May 19th 2008, 12:21 PM

Originally Posted by Mathstud28

Also $\forall{x}\in(0,1],\frac{1}{1-x^2}>\frac{1}{x^2}$

[spip]

So it implies that : $1-x^2<x^2 \Longleftrightarrow 2x^2>1$ , for $x \in [0,1]$ .

Now, tell me, am I misunderstanding you or are you just looking for something that would help you ?

**Mathstud28** · May 19th 2008, 01:07 PM

Originally Posted by Moo

So it implies that : $1-x^2<x^2 \Longleftrightarrow 2x^2>1$ , for $x \in [0,1]$ .

Now, tell me, am I misunderstanding you or are you just looking for something that would help you ?

I am not going to deny it this time. My calculator lied to me and I did not check it. My TI-89 has been acting up lately...sorry for the confusion

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