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

[Europio2]

Homework Statement

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

Homework Equations

\begin{equation} y = \int 1/(1-x)^2 \cdot dx \end{equation}

The Attempt at a Solution

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

What is wrong?

 

[SteamKing]

I know its easy, but I'm making a mistake somewhere that is making me crazy. I want to solve ∫1/(1-x)²dx

I use de sixth formula in this PDF, but it does not work http://integral-table.com/downloads/single-page-integral-table.pdf

I know the result is 1/(1-x), but using the formulo I get 1/(x-1)

What is wrong?

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

 

[Europio2]

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

 

[Fredrik]

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.

 

[Europio2]

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.

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

Look at this (this is correct).

\begin{equation} 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} \end{equation}

That's why I am confused.

 

[sushichan]

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

Look at this (this is correct).

\begin{equation} 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} \end{equation}

That's why I am confused.

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]

 

[Fredrik]

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

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

 

[Europio2]

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]

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

 

[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.

 

[sushichan]

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

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

 

[Fredrik]

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.

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

 

[Europio2]

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

¿Do you mean that?

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

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.

\begin{equation} 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} \end{equation}I'm missed up :(

 

[Fredrik]

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.

 

[Fredrik]

¿Do you mean that?

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.

 

[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.