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

1.	Evaluate:`int1/(x^4+1)dx`
Answer» Correct 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` `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`

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