求解答一道积分题~

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求解答一道积分题~

求解答一道积分题~
求解答一道积分题~

求解答一道积分题~
1/(x^4+1)=(√2/4x+1/2)/(x^2+√2x+1)+(-√2/4x+1/2)/(x^2-√2x+1)
∫[(√2/4)x+1/2]/(x^2+√2x+1)]dx
=(√2/8)[∫(2x+√2)/(x^2+√2x+1)dx+∫√2/(x^2+√2x+1)dx
=(√2/8)[ln(x^2+√2x+1)+2arctan(√2x+1)]+C
后半个式子方法相同
所以∫1/(x^4+1)dx
=(√2/8)[ln(x^2+√2x+1)-ln(x^2-√2x+1)+2arctan(√2x+1)+2arctan(√2x-1)]+C