Evaluate: `int1/(x^4+1)dx`

Answer

`(1)/(2sqrt(2))"tan"^(-1)((x^(2)-1)/(sqrt(2)x))-(1)/(4sqrt(2))"log"|(x^(2)-sqrt(2)x+1)/(x^(2)+xsqrt(2)+1)|+C`

Explanation

`I=int (1)/(x^(4)+1)dx=int((1)/(x^(2)))/(x^(2)+(1)/(x^(2)))dx`
`=(1)/(2)int((2)/(x^(2)))/(x^(2)+(1)/(x^(2)))dx`
`=(1)/(2)int((1+(1)/(x^(2)))/(x^(2)+(1)/(x^(2)))-(1-(1)/(x^(2)))/(x^(2)+(1)/(x^(2))))dx`
`=(1)/(2)int(1+(1)/(x^(2)))/(x^(2)+(1)/(x^(2)))dx-(1)/(2)int(1-(1)/(x^(2)))/(x^(2)+(1)/(x^(2)))dx`
`=(1)/(2)int(1+(1)/(x^(2)))/((x-(1)/(x))^(2)+2)dx-(1)/(2)int(1-(1)/(x^(2)))/((x+(1)/(x))^(2)-2)dx`
` "Putting "x-(1)/(x)=u" in first integral and "x+(1)/(x)=v " in second integral." `
we get
`I=(1)/(2)int(du)/(u^(2)+(sqrt(2))^(2))-(1)/(2)int(dv)/(v^(2)-(sqrt(2))^(2))`
`=(1)/(2sqrt(2))"tan"^(-1)((u)/(sqrt(2)))-(1)/(2)xx(1)/(2sqrt(2))"log"|(v-sqrt(2))/(v+sqrt(2))|+C`
`=(1)/(2sqrt(2))"tan"^(-1)((x-1//x)/(sqrt(2)))-(1)/(4sqrt(2))"log"|(x+1//x-sqrt(2))/(x+1//x+sqrt(2))|+C`
`=(1)/(2sqrt(2))"tan"^(-1)((x^(2)-1)/(sqrt(2)x))-(1)/(4sqrt(2))"log"|(x^(2)-sqrt(2)x+1)/(x^(2)+xsqrt(2)+1)|+C`