# Integrating 1/(1-x)^2 its making me crazy

1. Apr 22, 2015

### Europio2

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

I know its easy, but I'm making a mistake somewhere that is making me crazy. I want to solve $$y = \int 1/(1-x)^2 \cdot dx$$
I use de sixth formula in this PDF, but it does not work http://integral-table.com/downloads/single-page-integral-table.pdf

2. Relevant equations

$$y = \int 1/(1-x)^2 \cdot dx$$

3. The attempt at a solution

This is what I do.
$$y = \int 1/(1-x)^2 \cdot dx = \int (1-x)^{-2} \cdot dx = \frac{(1-x)^{-1}}{-1} = \frac{1}{x-1}$$

What is wrong?

Last edited: Apr 22, 2015
2. Apr 22, 2015

### SteamKing

Staff Emeritus
Well, you haven't shown your work, so how can anyone say what is wrong?

3. Apr 22, 2015

### Europio2

Sorry. This is what I do.
∫1/(1-x)^2dx = ∫(1-x)^-2dx = (1-x)^-1/-1 = 1/(x-1)

4. Apr 22, 2015

### Fredrik

Staff Emeritus
The formula you're trying to use says a+x, not a-x, so you should start with a rewrite that puts a plus sign in front of the x.

5. Apr 22, 2015

### Europio2

But that formula is valid for any other exponent I think.

Look at this (this is correct).

$$y = \int 1/(1-x)^5 \cdot dx = \int (1-x)^{-5} \cdot dx = \frac{(1-x)^{-4}}{-4} = \frac{1}{4(x-1)^4}$$

That's why I am confused.

6. Apr 22, 2015

### sushichan

Just a question from me: how do you enter an equation?

And notice that (1-x)^(-4)/(-4) does not equal to the two terms beside it. Basically you did two mistakes that made the sign change twice. [(1-x)^-4 = (x-1)^-4]

7. Apr 22, 2015

### Fredrik

Staff Emeritus
Choose "Help/How-To" on the "Info" menu in the upper right. Then click on "LaTeX Primer".

8. Apr 22, 2015

### Europio2

I do this -4*(1-x)^-4 = 4*(x-1)^-4, that's why I change the sign.

9. Apr 22, 2015

### epenguin

IYour first post is nearly right except there is a simple error in the last step, you have confused rules about straight minuses and minuses in an index it seems.

10. Apr 22, 2015

### sushichan

(1-x)^-4 = (-(x-1))^-4 = (-1)^-4 * (x-1)^-4 = (x-1)^-4

11. Apr 22, 2015

### Fredrik

Staff Emeritus
The last step is correct. It's the one before that (the one where he uses the formula from the pdf) that's wrong.

12. Apr 22, 2015

### Europio2

¿Do you mean that?

$$y = -\int -1/(1-x)^2 \cdot dx = -\int -(1-x)^{-2} \cdot dx = -\frac{(1-x)^{-1}}{-1} = \frac{1}{1-x}$$

But if I change the exponent to 5, as in the other post, and I do the same, I don't get the correct one.

$$y = -\int -1/(1-x)^5 \cdot dx = -\int -(1-x)^{-5} \cdot dx = -\frac{(1-x)^{-4}}{-4} = \frac{1}{4*(1-x)^4}$$

I'm missed up :(

13. Apr 22, 2015

### Fredrik

Staff Emeritus
Those extra two minus signs in the first calculation don't help. What I wanted you to do is to find a way to rewrite $\frac{1}{(1-x)^2}$ in the form $\frac{a}{(b+x)^2}$. You need a plus sign directly in front of the x before you can apply the formula.

Edit: My first version of this post contained an additional statement that was wrong. If you saw it, just ignore it. You should do a rewrite of the type described above at the start of both of these calculations. If you do, the results should be OK.

Last edited: Apr 22, 2015
14. Apr 22, 2015

### Fredrik

Staff Emeritus
I interpreted this as a question about the calculations in post #12, but since you were quoting me, I guess you were asking if I'm sure that the comment I had made earlier (the one you quoted) is right. I have checked it again, and I still say that you're using formula (6) wrong in post #1.

I will rephrase my comment about the rewrite you should do. If you want to be able to find primitive functions of $\frac{1}{(1-x)^n}$ for arbitrary n, you should rewrite $\int\frac{1}{(1-x)^n}dx$ in the form $b\int(x+a)^m dx$ and then use formula (6) to find $\int(x+a)^m dx$. In fact, I think you should leave n arbitrary (except for the requirement $n\neq 1$) and try this exact thing.

What you did in post #1 was to assume that you would end up with the right-hand side of formula (6) even though what you had didn't match the left-hand side. What you did in post #12 was to assume that the answer would be wrong by a factor of -1 regardless of what the exponent is. Both of these assumptions are wrong.

Last edited: Apr 22, 2015
15. Apr 22, 2015

### LCKurtz

I would add that, unless the OP was required to use a table for this problem for some reason, he shouldn't use the table in the first place. Just do a substitution $u=1-x$. He might even have gotten the correct answer the first time.