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This is the first dish \[\int \frac{x^2}{1-x^2}dx\] Apart from resolving that into partial fractions, what is a better way to do it?

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let x=sinu \[\int \frac{x^2}{1-x^2}dx = \int\frac{sin^2ucosu}{cos^2u}du=\int(sec^2u-cosu)du=tanu-sinu+C\]?
Last line there: tanu -sinu +C \[=\frac{x}{x^2+1}-x+C\] What am I doing here :S
\[\frac{sin^2ucosu}{cos^2u}=\frac{(1-cos^2u)cosu}{cos^2u} = sec^2u-cosu\]

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Other answers:

tan u is x/ sqrt(1-x^2)
Oh.. My mistake! \[=\frac{x}{1-x^2} -x+C\]
Alright, let's go to the main dish :S
and its not sec^2 , check again
secu! I'd better do it all over again!!! \[\int \frac{x^2}{1-x^2}dx \]\[= \int\frac{sin^2ucosu}{cos^2u}du\]\[=\int(secu-cosu)du\]\[=\ln|secu+tanu|-sinu+C\]
not a better way......
I know, I just try whatever I can.. Is there a better way then?
i don't think,partial is best....
\[\int \frac{x^2}{1-x^2}dx\]\[ = \int (\frac{1}{2(x+1)}-\frac{1}{2(x-1)}-1)dx\]\[=\frac{1}{2} ln|x+1| -\frac{1}{2}ln|x-1|-x+C\]\[=\frac{1}{2}ln|\frac{x+1}{x-1}|-x+C\]
yup, thats correct.....

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