`int(x^2)/(1-x^4)dx`

Answer

N/a

Explanation

`I = int(x^(2))/(1-x^(4))dx`
` = int((1/2+(x^(2))/(2)-1/2+(x^(2))/(2)))/((1-x^(2))(1+x^(2)))dx` , `[:'a^(2)-b^(2)=(a+b)(a-b)]`
`= int(1/2(1+x^(2))-1/2(1-x^(2)))/((1-x^(2))(1+x^(2)))dx`
`= int(1/2(1+x^(2)))/((1+x^(2))(1+x^(2)))dx-1/2int((1-x^(2)))/((1-x^(2))(1+x^(2)))dx`
` = 1/2 int(1)/(1-x^(2))dx - 1/2 int(1)/(1+x^(2))dx = 1/(2).(1)/(2)log|(1+x)/(1-x)|+C_(1)-1/2tan^(-1)x+C_(2)`
`= 1/4 log|(1+x)/(1-x)|-1/2tan^(-1)x+C , [:' C = C_(1)+ C_(2)]`