1. ## Easy integral!!

Solve the following integral

2. Originally Posted by matty888
Solve the following integral
Use a partial fraction decomposition.

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

4. 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 .....

5. 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..

6. 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??

7. 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.

8. 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

9. 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

10. Originally Posted by Mathstud28
Also $\forall{x}\in(0,1],\frac{1}{1-x^2}>\frac{1}{x^2}$

[spip]
So it implies that : $1-x^21$, for $x \in [0,1]$.

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

11. Originally Posted by Moo
So it implies that : $1-x^21$, 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