How to Solve Improper Fractions

[yoleven]

Homework Statement

[tex]\int \frac{x^{2}}{x^{2}-1}[/tex]

Homework Equations

The Attempt at a Solution

I divide bottom into the top because the degree is the same and get...
[tex]\int 1-\frac{1}{x^{2}+1}[/tex]

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

 

[statdad]

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

 

[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.

 

[The Dagda]

Never mind

 

[yoleven]

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

 

[HallsofIvy]

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

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

 

[The Dagda]

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

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

 

[uzairi22]

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

 

[n!kofeyn]

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

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.

 

[HallsofIvy]

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

First, please do not "hijack" someone else's thread for a completely new question. Start your own thread using the "new post" button.

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

 

[n!kofeyn]

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

First, please do not "hijack" someone else's thread for a completely new question. Start your own thread using the "new post" button.

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

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.

 

[n!kofeyn]

Homework Statement

[tex]\int \frac{x^{2}}{x^{2}-1}[/tex]

The Attempt at a Solution

I divide bottom into the top because the degree is the same and get...
[tex]\int 1-\frac{1}{x^{2}+1}[/tex]

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

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:
[tex]\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[/tex]
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.