Integrate a fraction with x function on top and polynomial on bottom

Full question:

Evaluate the following integral(s) (I'm posting the first one):

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

I have no idea how to do it.. I tried using partial fractions and got something horrible like

ln[(x+1)^{-0.5} (x-1)^{1.5}] + lnk

Re: Integrate a fraction with x function on top and polynomial on bottom

Quote:

Originally Posted by

**Mukilab** Evaluate the following integral(s) (I'm posting the first one):

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

ln[(x+1)^{-0.5} (x-1)^{1.5}] + lnk

Look at this calculation.

Re: Integrate a fraction with x function on top and polynomial on bottom

Ok so that's partial fractions but I wouldn't know where to go from there.

How do I integrate the (x+1)^-1 or (x-1)^-1?

Re: Integrate a fraction with x function on top and polynomial on bottom

Quote:

Originally Posted by

**Mukilab** Ok so that's partial fractions but I wouldn't know where to go from there.

How do I integrate the (x+1)^-1 or (x-1)^-1?

Use a substitution:

$\displaystyle \int \frac{dx}{x + 1}$

Let u = x + 1 implies du = dx, etc.

$\displaystyle \int \frac{dx}{x + 1} = \int \frac{du}{u}$

Can you finish it now?

-Dan

Re: Integrate a fraction with x function on top and polynomial on bottom

Thanks for your help topsquark.

Is the dx you are using what is usually at the end of an integration (you are integrating with respect to d) or is it just a constant? If it is the former, I had no idea you're allowed to put it into the equation that you are integrating.

Assuming it is just a constant, your du/u would give you d. Integrating this would give you x.

Therefore for my integral above, would this be correct?:

$\displaystyle \int \frac {x+2}{(x+1)(x-1)} dx$ = $\displaystyle \frac {3}{2} \int \frac {1}{x-1} dx - \frac {1}{2} \int \frac {1}{x+1} dx$ = $\displaystyle 1$

Re: Integrate a fraction with x function on top and polynomial on bottom

Quote:

Originally Posted by

**Mukilab** Thanks for your help topsquark.

Is the dx you are using what is usually at the end of an integration (you are integrating with respect to d) or is it just a constant? If it is the former, I had no idea you're allowed to put it into the equation that you are integrating.

Assuming it is just a constant, yofur du/u would give you d. Integrating this would give you x.

Therefore for my integral above, would this be correct?:

$\displaystyle \int \frac {x+2}{(x+1)(x-1)} dx$ = $\displaystyle \frac {3}{2} \int \frac {1}{x-1} dx - \frac {1}{2} \int \frac {1}{x+1} dx$ = $\displaystyle 1$

The "dx" gets moved around sometimes. You typically see it in one of three places:

$\displaystyle \int dx (...) \text{ } \int \frac{dx}{...} \text{ } \int ... dx$

It all means the same thing.

The partial fraction decomposition is good, but the RHS is not correct. The fractions don't add up to 1 in general. The idea is to break the denominator into linear factors (if possible) then integrating them to get a ln function.

-Dan

Re: Integrate a fraction with x function on top and polynomial on bottom

Quote:

Originally Posted by

**topsquark** The "dx" gets moved around sometimes. You typically see it in one of three places:

$\displaystyle \int dx (...) \text{ } \int \frac{dx}{...} \text{ } \int ... dx$

It all means the same thing.

The partial fraction decomposition is good, but the RHS is not correct. The fractions don't add up to 1 in general. The idea is to break the denominator into linear factors (if possible) then integrating them to get a ln function.

-Dan

Thank you, sorry that was a bit silly of me.

From $\displaystyle \int \frac {x+2}{(x+1)(x-1)} dx$ = $\displaystyle \frac {3}{2} \int \frac {1}{x-1} dx - \frac {1}{2} \int \frac {1}{x+1} dx$ = $\displaystyle 1$

The answer would be

$\displaystyle \frac {3}{2} ln(x-1) - \frac {1}{2} ln(x+1) + ln(k) $

Is that correct?

It can be simplified into $\displaystyle ln \frac{k(x-1)^{\frac{3}{2}}}{(x+1)^\frac{1}{2}} $

I still don't see how you would use the u=x+1 substitution though. You would have to integrate a constant d.... with respect to...? Assuming with respect to x you would integrate it as x. Giving the answer to be 3/2(x) - 1/2(x) + c which isn't the same as the ln one.

Please could you explain the process for substituting u in and then solving it? I've never seen it done before nor have I done it myself. I've *just* learned about integrating to get ln a few days ago in class.

Re: Integrate a fraction with x function on top and polynomial on bottom

There is NO "constant d! That was what topsquark told you and you said you understood.

$\displaystyle \int \frac{du}{u}= \int\frac{1}{u}du= ln(u)+ C$. Whoever gave you this problem clearly expects you to know that.

Re: Integrate a fraction with x function on top and polynomial on bottom

At no point did topsquark say d is not a constant :(

Thanks for the formula, I had no idea about it. The question is from the physics aptitude test for oxford university which can feature stuff that can at times be far beyond my syllabus or not taught in schools such as lunar phases... or the above (although I'm guessing it will be taught at some stage in my education).

Thanks again :)

Re: Integrate a fraction with x function on top and polynomial on bottom

Quote:

Originally Posted by

**Mukilab** At no point did topsquark say d is not a constant :(

On the other hand I would have thought that anyone posting in the Calculus forum and integrating rational polynomials would be aware that the "d" in "dx" did not denote a constant.

If this is news to you I'd recommend a good refresher.

-Dan

Re: Integrate a fraction with x function on top and polynomial on bottom

Quote:

Originally Posted by

**topsquark** On the other hand I would have thought that anyone posting in the Calculus forum and integrating rational polynomials would be aware that the "d" in "dx" did not denote a constant.

If this is news to you I'd recommend a good refresher.

-Dan

I wanted to post in pre-calculus since I'm not at University yet but the rules said no integrals allowed there...

Re: Integrate a fraction with x function on top and polynomial on bottom

Quote:

Originally Posted by

**HallsofIvy** There is NO "constant d! That was what topsquark told you and you said you understood.

$\displaystyle \int \frac{du}{u}= \int\frac{1}{u}du= ln(u)+ C$. Whoever gave you this problem clearly expects you to know that.

Actually it's $\displaystyle \displaystyle \begin{align*} \int{ \frac{du}{u} } = \int{ \frac{1}{u}\,du} = \ln{ |u | } + C \end{align*}$

Quote:

Originally Posted by

**Mukilab** I wanted to post in pre-calculus since I'm not at University yet but the rules said no integrals allowed there...

Why on Earth would you be asked to integrate something if "no integrals are allowed"?

Re: Integrate a fraction with x function on top and polynomial on bottom

Quote:

Originally Posted by

**Prove It** Why on Earth would you be asked to integrate something if "no integrals are allowed"?

I think Mukilab was trying to say that (s)he isn't in a college Calculus class and the Forum doesn't allow for integral questions in the Pre-Calculus forum.

Still, if it's an integration problem, it's Calculus and you posted in the correct forum. My point still stands...If you are this far along and having that basic a problem you need to consult with your instructor.

-Dan

Re: Integrate a fraction with x function on top and polynomial on bottom

Thanks guys, is there any different between putting ln(k) or C for the constant?

Re: Integrate a fraction with x function on top and polynomial on bottom

Quote:

Originally Posted by

**Mukilab** Thanks guys, is there any different between putting ln(k) or C for the constant?

Not really. But it is usually "nicer" to have the ln(k) form to write the answer a bit more neatly.

-Dan