# 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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus

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.