Evaluate: `int(x^2+4)/(x^4+16)dx`

Explanation

`I=int(x^(2)+4)/(x^(4)+16)dx`
`=int(1+(4)/(x^(2)))/(x^(2)+(16)/(x^(2)))dx`
`=int(1+(4)/(x^(2)))/(x^(2)+((4)/(x))^(2)-8+8)dx`
`=int(1+(4)/x^(2))/((x-(4)/(x))^(2)+8)dx`
Let `x-(4)/(x)=t. " Then " d(x-(4)/(x))=dt " or " (1+(4)/(x^(2)))dx=dt.`
Therefore,
`I=int(dt)/(t^(2)+(2sqrt(2))^(2))=(1)/(2sqrt(2))tan^(-1)((t)/(2sqrt(2)))+C`
`=(1)/(2sqrt(2))tan^(-1)((x-(4)/(x))/(2sqrt(2)))+C`
`=(1)/(2sqrt(2))tan^(-1)((x^(2)-4)/(2x sqrt(2)))+C`