Question

what method of integration would I use to solve ((x^2)/(1-x^2))dx?

Answer 1

\[(1-x^2)=(x-1)(x+1)\]now use partial fraction decomposition.

this is the first thing I notice but might not be the best rout...give it a shot

Answer 2

i was thinking partial fractions? but idk how you would split them up

Answer 3

Can do u-sub. u = 1-x^2, meaning x^2 = 1-u, which would give you (1-u)/u. Then just split that into two fractions and integrate.

Answer 4

hahah wow that was a blonde moment...thank you:D

Answer 5

\[(1-x^2)=(1-x)(1+x)\]

Answer 6

had to fix my mistake:)

Answer 7

Partial decomp just seems a lot longer to do when you can u-sub xD

Answer 8

We have to borrow one of the x's for the differential du.
So let's be careful doing it this way! :o
\[\Large \int\limits \frac{x}{1-x^2}(x\;dx)\]\[\Large u=1-x^2 \qquad\to\qquad -\frac{1}{2}du=(x\;dx)\]\[\Large \int\limits \frac{\sqrt{1-u}}{u}\left(-\frac{1}{2}du\right)\]

Answer 9

Its x^2 on top, zep xD

Answer 10

he moved that x

Answer 11

\[\Large \int\limits \frac{x^2}{1-x^2}dx \quad=\quad \int\limits \frac{x}{1-x^2}(x\;dx)\]

Answer 12

x*xdx = x^2dx

Answer 13

We need one of the x's for the dx. See what happens? :(

Answer 14

didn't even notice obviously. Im dumb, lol.

Answer 15

Hmm yah, my vote is for Partial Fraction Decomp :D heh

Answer 16

Guess that doesn't make it easier :/

Answer 17

Ya not so much I guess :D
I mean if you remember this integral:\[\Large \int\limits \frac{1}{1-x^2}\;dx \quad=\quad \tanh^{-1} x\]
Then fine.. :P but most of us don't, lol.

Answer 18

Did you ever figure this problem out @megannicole51 ? :o

Answer 19

yeah i know how to do those now thank you:)