ज्ञात करें `int (1+x^2)/(1+x^4)dx`

Explanation

`I =int (1+x^2)/(1+x^4)dx = int (1/x^2+1)/(1/x^2+x^2)dx = int (1+1/x^2)/((x-1/x)^2 +2)` [ जहां `1+1/x^2=(x-1/x)` का `d.c.` ]
`z=x-1/x` रखे, तो `dx=(1+1/x^2)dx`
अब, `I = int (dz)/(z^2+2) = int (dz)/((sqrt2)^2+z^2) =1/sqrt2 tan^(-1)z/sqrt2 +c`
`=1/sqrt2 tan^(-1) ((x-1/x)/sqrt2)+c=1/sqrt2 tan^(-1) ((x^2-1)/(sqrt2x))+c`