Integration help needed

1. Dec 8, 2007

erjkism

1. The problem statement, all variables and given/known data

$$\int(\frac{x^{2}+1 dx}{x-1}^{2})$$

i tried to get 'dx' into the integral but it didnt work out

2. Relevant equations

3. The attempt at a solution

i have tried squaring the top and bottom, which gave {x$$^{4}$$ + 2x$$^{2}$$ +1}/{x^{2} + 2x +1}

i dont know if that was a good idea, but i was stumped. I cant seem to get anything to cancel out

2. Dec 8, 2007

rocomath

hm, i'm not quite sure what your problem is

$$\int(\frac{x^2 +1}{x-1})^2 dx$$

is this correct?

3. Dec 8, 2007

erjkism

yea that's it.

4. Dec 8, 2007

rocomath

What integration technique are you learning at the moment? I'm just wondering so I can show you with that method.

btw, it should be x^2 - 2x + 1 in the denominator.

5. Dec 8, 2007

erjkism

oh yeah you're right.
well, so far we've learned the fundamental theorem of calculus and ordinary U substitutions. i havent gotten to the point of using special formulas for anything yet.

6. Dec 8, 2007

rocomath

Wow. This problem is giving me trouble, I feel ashamed!!!

i have

$$\int\frac{(x^2 +1)^2}{(x^2 +1) -2x}dx$$

but it's not leading me anywhere, argh.

7. Dec 8, 2007

malty

Hey, join the club!!

I got it down to $$1+ \int{\frac{x+1}{x-1}}dx$$ is there anything that can be done here... ?

...wait I think i forgot to square it :(

Yup

Last edited: Dec 8, 2007
8. Dec 8, 2007

Dick

Don't feel ashamed. Given what the OP said about the level of the course, this problem should never have been given before teaching partial fractions.

9. Dec 8, 2007

Mystic998

Eh, it's doable with u-substitution. I don't want to just hand out the answer, but square and do polynomial division. Then you get a quadratic that's easy to integrate and something involving a u substitution.

10. Dec 8, 2007

rocomath

Thanks! I was feeling kinda crappy, lol. I'm actually in Calculus 2 so I should be able to do this, but it wasn't working out for me with simpler techniques.

Well, I didn't think that. Uh!!! That should definitely work.

11. Dec 8, 2007

Dick

Good point. That's the other way to do it.

12. Dec 8, 2007

rocomath

SOLVED! Now I''m happy :-] Thanks.

13. Dec 8, 2007

erjkism

when i used polynomial division i got big mess:

(x^2)-2x+7+ [(12x-6)/((x^2)-2x+1)]

so i have to use a U sub. now?

Last edited: Dec 8, 2007
14. Dec 8, 2007

rocomath

from my polynomial division if i remember correctly, was different. let me re-do it real fast.

15. Dec 8, 2007

rocomath

Polynomial division, let's do this part first so our Integration will work out.

$$\int\frac{x^4 +2x^2 +1}{x^2 -2x +1}dx=\int(x^2 +2x +5)+\frac{8x-4}{x^2 - 2x+1}dx$$

Last edited: Dec 8, 2007
16. Dec 8, 2007

erjkism

k isee that lol im an idiot. i havent done it in a while. i get it. but what about the U sub?

17. Dec 8, 2007

rocomath

k, I feel good about my Polynomial now.

so the u-sub ...

$$\int(x^2 +2x +5)+\frac{8x-4}{x^2 - 2x+1}dx$$

now splitting it into two Integrals

$$\int(x^2 +2x +5)dx+\int\frac{8x-4}{x^2 -2x +1}dx$$

so now doing a u-substitution only for the second Integral

$$\int\frac{4(2x-1)}{x^2 -2x +1}dx$$

$$u=x^2 -2x +1$$

now just take the derivative and you will need to mess around with the numerator a little more to get your du in your Integral.

Last edited: Dec 8, 2007
18. Dec 8, 2007

erjkism

thanks man i really appreciate it. sorry for making you type all that up lol, but i got a calc final coming up

19. Dec 8, 2007

rocomath

No problem! You'll do fine :-] And the typing was no biggy, it was mainly copy/paste :-]

20. Dec 9, 2007

coomast

Another possibility is to use the substitution:

$$u=x-1$$

On the last integral rewritten as:

$$4\int\frac{2x-1}{(x-1)^2}dx$$

I find this a bit easier. The different methods can be compared, all are equally valid.

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