Integral of improper fraction

1. Feb 11, 2009

yoleven

1. The problem statement, all variables and given/known data
$$\int \frac{x^{2}}{x^{2}-1}$$

2. Relevant equations

3. The attempt at a solution
I divide bottom into the top because the degree is the same and get...
$$\int 1-\frac{1}{x^{2}+1}$$

My question is, from here, do I have to recognize the tan$$^{-1}$$
or is it also correct to put
x-ln $$\left| x^{2} + 1\right|$$ + c ?

2. Feb 11, 2009

Why would you use the logarithm when it isn't correct?

3. Feb 11, 2009

HallsofIvy

Surely you know that you can't just "ignore" functions like that. The integral of 1/f(x), in general, has nothing to do with the integral of 1/x. Yes, you really have to be able to "recognize" the basic integrals.

4. Feb 11, 2009

The Dagda

Never mind

Last edited: Feb 11, 2009
5. Feb 11, 2009

yoleven

I never claimed to be overly bright. I asked because I was confused.
Thanks for responding.

6. Feb 11, 2009

HallsofIvy

Join the club! Most of us here are not overly bright and often confused.:tongue2:

7. Feb 12, 2009

The Dagda

You only have to check the comment in my edit to realise that is true, my excuse is that I'd not long woken up and the government, ahem *cough*.

8. Apr 3, 2009

uzairi22

can u help me with the integral of (2x^3)/(x-1)

9. Apr 3, 2009

n!kofeyn

Whenever you have a troublesome integral, it always helps to start with trying u-substitution or integration by parts. Which technique do you think you should use here? Sometimes you just have to pick a method and try it out. If it fails, see if you can modify it so that it works. A further hint can be provided if you try these methods out and are still stuck.

10. Apr 3, 2009

HallsofIvy

Now, first divide $2x^3+ 0x^2+ 0x+ 0$ by x-1 to get a quadratic polynomial plus something of the form A/(x-1). Integrate that.

11. Apr 3, 2009

n!kofeyn

I at first disregarded dividing by [itex]x-1[/tex], but it works out quickly (say with synthetic division or long division). If you have an idea, work it out! Letting [itex]u=x-1[/tex] works out fine, but just requires more algebra. Both methods give the same answer.

Thank you for the forum advice HallsofIvy.

12. Apr 3, 2009

n!kofeyn

Back to the original post. You have a few signs wrong (add everything back together and you don't get what you started with). After using long division to divide [itex]x^2-1[/tex] into [itex]x^2[/tex] (be sure that you can do this) you get:
$$\int \frac{x^{2}}{x^{2}-1} \,dx = \int \left( 1 + \frac{1}{x^2-1} \right) \,dx = \int 1 \,dx + \int \frac{1}{(x-1)(x+1)} \,dx = x + C + \int \frac{1}{(x-1)(x+1)} \,dx$$
Now you expand the fraction on the far right using partial fractions. Then you will get integrals that can be evaluated using the natural log.