Integrate f(x) = \frac{1}{(x-1)(x-2)} (1<x<2)

[Kawakaze]

Homework Statement

Determine indefinite integral of

$$f(x) = \frac{1}{(x-1)(x-2)}$$ (1 < x < 2)

The Attempt at a Solution

Use standard integral

$$\frac{1}{(x - a)(x - b)} = \frac{1}{a - b}ln\frac{x - a}{x - b}$$

This doesn't give me the same answer as wolfram alpha or mathcad

Should instead treat this as a compound function?

$$f(x) = \frac{1}{(x - 1)}\frac{1}{(x - 2)}$$

 

[Gib Z]

I suspect that wolfram alpha and mathcad gave you answers in terms of inverse hyperbolic trig functions. If so, look it up on google, inverse hyperbolic trig functions are expressible in terms of logs.

 

[Kawakaze]

Hi Gib

Thanks for the reply, unfortunately it may as well have been in russian. =)

My method using the standard integrals gives

$$-ln\frac{x - 1}{x - 2}$$

Which I can't simplify any more really? Wolfram alpha and mathcad both give

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

 

[Gib Z]

Your answers are eqivalent, using the log law: log(a/b) = log a - log b. Also, don't forget the absolute value sign your standard integrals should have.

 

[Kawakaze]

Thanks Gib, that's good news, and I wasnt aware of the equivalency either =)

What do you mean absolute sign my standard integrals should have? I can't see an incorrect sign, please explain.

Lastly, should there be a +C at the end of this or not?

 

[Gib Z]

$\int \frac{1}{x} dx = \log |x| + C$, not just log x. Yes, there's always a + C at the end, its often omitted as it becomes assumed. Anyway, you need the |x| to ensure the term being log-ed is always positive.

So instead of $- \log \left( \frac{x-1}{x-2} \right) + C$, you should have had $- \log \left| \frac{x-1}{x-2} \right| + C = - \log \left( \frac{x-1}{2-x} \right) + C$, since 1 < x < 2 so $|x-1| = x-1$ and $|x-2| = 2-x$.