# Problem on integration

1. Feb 28, 2005

### trap

can anyone help me with this,

$$\int \frac{\ x^2}{1+x^2} dx$$

thanks.

I've tried with integration by parts but does not work..

2. Mar 1, 2005

### Moo Of Doom

Hmm... I don't quite remember how to use partial fractions for that... but I remember there was a way (at least with a similar problem). You could try

$$\int{\frac{x^2}{1+x^2}dx}=\int{\frac{x^2}{(x-i)(x+i)}dx}$$

and use partial fractions from there (but it gets a bit "hairy," as my math teacher always says).

3. Mar 1, 2005

### daveyp225

Did you try partial fractions?

4. Mar 1, 2005

### Townsend

Partial fractions work only for proper fractions. This is not a proper fraction in the sense that the degree of the numerator is the same as the degree of the denominator. So to fix this little problem you need to use long division after which you get.

$$1-\frac{1}{x^2+1}$$

Which is now easy to integrate.

Regards

Last edited: Mar 1, 2005
5. Mar 1, 2005

### Muzza

Use x^2/(1 + x^2) = (x^2 + 1 - 1)/(1 + x^2) = 1 - 1/(1 + x^2) and integrate that instead...

6. Mar 1, 2005

### dextercioby

In this case partial fractions would lead indeed to a decomposition containing complex numbers.Therefore the method is not good.

Daniel.

7. Mar 1, 2005

### Jameson

I did this on my calculator and got a simple answer. I think you can do integration by parts. What have you tried?

8. Mar 1, 2005

### Jameson

Integration by parts isn't necessary. Townsend clearly explains the best way to do this.

$$\int\frac{x^2}{x^2+1} dx = \int\frac{x^2+1}{x^2+1}-\frac{1}{x^2+1}dx$$

Hint for the second part: trig inverses.

9. Mar 1, 2005

### trap

thank you everyone for the help, i understand now