Answer 1

`int x^2/(x^4+x^2-2) dx`

`=int x^2/((x2-1)(x^2+2)) dx`

`therefore x^2/((x2-1)(x^2+2)) =z/((z-1)(z+2))`

Using partial fraction, we have

`z/((z-1)(z+2))=A/(z-1)+B/(z+2)`

When z=1, A=1/3

When z=−2, B=2/3

`therefore int x^2/((x2-1)(x^2+2)) dx`

`=int (1/3)/(x^2-1^2) dx +int (2/3 )/(x^2+2)dx`

`=1/2 int 1/(x^2-1^2) dx +2/3 int (1 )/(x^2+2)dx`

`=1/3 xx 1/2 log |(x-1)/(x+1)|+2/3 xx 1/sqrt2 tan^-1 (x/sqrt2)+c`

`=1/6 log|(x-1)/(x+1)|+sqrt2/3 tan^-1(x/sqrt2)+c`