This Integral is Giving Me Trouble...

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April 18th 2008, 03:06 PM
Kalter Tod

This Integral is Giving Me Trouble...

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

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

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.
April 18th 2008, 03:13 PM
janvdl

Quote:

Originally Posted by Kalter Tod

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

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

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.

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

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

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

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

$= x - arctan(x) + C$
April 18th 2008, 06:52 PM
Mathstud28

Quote:

Originally Posted by Kalter Tod

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

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

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