# Thread: This Integral is Giving Me Trouble...

1. ## This Integral is Giving Me Trouble...

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

I set up a substitution and made $\displaystyle u=x^2+1$ so I could rewrite the integral as $\displaystyle \frac{u-1}{u}$ but my TI-89 is giving me the answer as $\displaystyle x-\arctan{x}$

I was thinking of using a trig-substitution, but this is for a friend of mine's homework, and he hasn't learned those yet. I know it's something simple that I'm missing, but I can't figure it out.

2. Originally Posted by Kalter Tod
$\displaystyle \int{\frac{x^2}{x^2+1}}$

I set up a substitution and made $\displaystyle u=x^2+1$ so I could rewrite the integral as $\displaystyle \frac{u-1}{u}$ but my TI-89 is giving me the answer as $\displaystyle x-\arctan{x}$

I was thinking of using a trig-substitution, but this is for a friend of mine's homework, and he hasn't learned those yet. I know it's something simple that I'm missing, but I can't figure it out.
$\displaystyle \int \frac{x^2}{x^2 + 1}dx$

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

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

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

$\displaystyle = x - arctan(x) + C$

3. Originally Posted by Kalter Tod
$\displaystyle \int{\frac{x^2}{x^2+1}}$

I set up a substitution and made $\displaystyle u=x^2+1$ so I could rewrite the integral as $\displaystyle \frac{u-1}{u}$ but my TI-89 is giving me the answer as $\displaystyle x-\arctan{x}$

I was thinking of using a trig-substitution, but this is for a friend of mine's homework, and he hasn't learned those yet. I know it's something simple that I'm missing, but I can't figure it out.
Generally when the numerator contains a power that is equal or one degree greater than the denominator you use polynomial division (or this technique that jandvl uses which is what I do) it is the best way to do these