Integration by Partial Fractions

[kdinser]

I'm making a small mistake somewhere, but I can't seem to find it.

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

taking the partial fractions

$$1=A(1-2x)+B(x-1)$$

$$A=-1, B=-2$$

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

Integrating by substitution, this is what I'm getting

$$-ln(x-1)+ln(1-2x)+C$$

The correct answer is

$$-ln(x-1)+ln(2x-1)+C$$

I think I'm making some kind of algebraic sign mistake when taking the second integral, but I just can't find it. I'll go through my solution step by step here and if someone could point out the mistake I would appreciate it.

$$\int\frac{-2}{1-2x}dx$$

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

$$u=1-2x$$ $$\frac{du}{dx}=-2$$

$$-2\int\frac{dx}{-2u}$$

$$\int\frac{dx}{u}$$

$$ln(u)+C$$

$$ln(1-2x)+C$$

Somewhere along the way, I'm missing the spot where I'm supposed to factor out a -1 from the denominator, but where?

 

[arildno]

Your mistake is rather subtle, it is basically that you think the anti-derivative of 1/x is ln(x); the right expression is ln(|x|) (that is, an absolute value sign must be included)

To give an example:
Look at 1/(x-1).
On the face of it, we would conclude that the antiderivative is ln(x-1).

But, you could always write:
1/(x-1)=(-1)/(1-x) (Agreed?)
But if you integrate the right-hand side, you'll seemingly end up with ln(1-x)

As you can see, all this makes sense, only if you realize that the correct anti-derivative is ln(|x-1|)

 

[kdinser]

Thanks for the quick reply, that explains why it wasn't obvious.

 

[kdinser]

One more thing.

Just so I can make sure I understand it. That means I would be free to use either solution?

The reason I ask is, this integral was part of a larger diff eq problem where I needed ln(2x-1) to verify the solution of a diff eq.

My TI-89 gives the solution ln(|2x-1|), I can't believe that I didn't notice the || signs the first time I looked at it . Is there a reason that ln|2x-1| is more correct then ln|1-2x|?

 

[Chen]

ln|2x-1| is identical in every way to ln|1-2x|, I see no reason why you wouldn't be able to use either expression.

It just so happens I made the exact same mistake a couple of days, and spent hours trying to figure out why Mathematica gave me an answer of ln|(x-1)/(x+1)| while I was getting ln|(1-x)/(1+x)|...

 

[arildno]

Is there a reason that ln|2x-1| is more correct then ln|1-2x|?

Absolutely NONE whatsoever, since we always have: |a|=|-a| for every choice of a.

So, as long as you keep the absolute value sign in there, you're free to use either expression (since they are equal)

 

[kdinser]

Great, thanks guys.