Question

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?

Answer 1

let x=sinu
\[\int \frac{x^2}{1-x^2}dx = \int\frac{sin^2ucosu}{cos^2u}du=\int(sec^2u-cosu)du=tanu-sinu+C\]?

Answer 2

Last line there: tanu -sinu +C
\[=\frac{x}{x^2+1}-x+C\]
What am I doing here :S

Answer 3

\[\frac{sin^2ucosu}{cos^2u}=\frac{(1-cos^2u)cosu}{cos^2u} = sec^2u-cosu\]

Answer 4

tan u is x/ sqrt(1-x^2)

Answer 5

Oh.. My mistake!
\[=\frac{x}{1-x^2} -x+C\]

Answer 6

Alright, let's go to the main dish :S

Answer 7

and its not sec^2 , check again

Answer 8

sec-cos

Answer 9

secu!
I'd better do it all over again!!!
\[\int \frac{x^2}{1-x^2}dx \]\[= \int\frac{sin^2ucosu}{cos^2u}du\]\[=\int(secu-cosu)du\]\[=\ln|secu+tanu|-sinu+C\]

Answer 10

not a better way......

Answer 11

I know, I just try whatever I can.. Is there a better way then?

Answer 12

i don't think,partial is best....

Answer 13

\[\int \frac{x^2}{1-x^2}dx\]\[ = \int (\frac{1}{2(x+1)}-\frac{1}{2(x-1)}-1)dx\]\[=\frac{1}{2} ln|x+1| -\frac{1}{2}ln|x-1|-x+C\]\[=\frac{1}{2}ln|\frac{x+1}{x-1}|-x+C\]

Answer 14

yup, thats correct.....