Question

Evaluate:

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

Answer 1

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

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

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

Using partial fraction

`x^/((x+1)(x-1)(x^2+2))=A/(x-1)+B/(x+1)+(Cx+D)/(x^2+2)`

`=(A(x+1)(x^2+2)+B(x-1)(x^2+2)+(Cx+D)(x+1)(x-1))/((x+1)(x-1)(x^2+2))`

Equating the coefficients from both the numerators we get,

A + B + C = 0........(1)
A - B + D = 1........(2)
2A + 2B - C = 0........(3)
2A - 2B - D= 0........(4)

Solving the above equations we get,

`A=1/6, B=-1/6, C=0, D=2/3`

Our Integral becomes

`intx^/((x+1)(x-1)(x^2+2))dx=1/(6(x-1))-1/(6(x+1))+2/(3(x^2+2))dx`

`=1/6log(x-1)-1/6log(x+1)+2/3xx1/sqrt2 tan^-1 (x/sqrt2)+C`

`=1/6[log(x-1)-log(x+1)+2sqrt2tan^-1 (x/sqrt2)]+C`