# Homework Help: 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.