Integrate this!

**freestar** · January 14th, 2012, 03:39 PM

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

Tried both u substitution letting u = 1+x^2, and u = atan(x), neither worked. Tried using integration by parts as well. Any suggestions in the right direction?

**e^(i*pi)** · January 14th, 2012, 04:11 PM

Originally Posted by freestar

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

Tried both u substitution letting u = 1+x^2, and u = atan(x), neither worked. Tried using integration by parts as well. Any suggestions in the right direction?

$\dfrac{x^2}{1+x^2} = \dfrac{x^2+1-1}{1+x^2} = \dfrac{x^2+1}{1+x^2} - \dfrac{1}{1+x^2} = 1 - \dfrac{1}{1+x^2}$

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

The second integral is a standard integral according to Wolfram (are you "allowed" to know this result or do you have to show/derive it?)

**Tikoloshe** · January 14th, 2012, 04:25 PM

You can also use the substitution $\tan u=x$ directly in the first integral. Then keep in mind that $1+\tan^2u=\sec^2u$ .

**freestar** · January 14th, 2012, 04:32 PM

Thank you sir! I am allowed to know this result. I'll keep this trick in mind for the future. Much appreciated.

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