## RolyPoly 3 years ago 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?

1. RolyPoly

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

2. RolyPoly

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

3. RolyPoly

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

4. hartnn

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

5. RolyPoly

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

6. RolyPoly

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

7. hartnn

and its not sec^2 , check again

8. hartnn

sec-cos

9. RolyPoly

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$

10. hartnn

not a better way......

11. RolyPoly

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

12. hartnn

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

13. RolyPoly

$\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$

14. hartnn

yup, thats correct.....