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这类题目(求分式的不定积分)首先要进行有理分式分解:首先因式分解分母:多项式的根为±(√2/2+√2/2i)和±(-√2/2+√2/2i)所以x^4+1=(x^2+√2x+0.5)(x^2-√2x+0.5)=(x+√2/2)^2(x-√2/2)^2(因式定理)
(x^2+1)/(x^4+1)=A/(x+√2/2)^2+B/(x+√2/2)+C/(x-√2/2)^2+D/(x-√2/2),解方程组即可得到A=3/4 B=√2/4 C=3/4 D=-√2/4
积分(x^2+1)/(x^4+1)dx=积分(3/4)/(x+√2/2)^2+(√2/4)/(x+√2/2)d(x+√2/2) +(3/4)/(x-√2/2)^2-(√2/4)/(x-√2/2)d(x-√2/2)=√2/4ln(x+√2/2)-√2/4ln(x-√2/2)-(3/4)/(x+√2/2)-(3/4)/(x+√2/2)+C
(x^2+1)/(x^4+1)=A/(x+√2/2)^2+B/(x+√2/2)+C/(x-√2/2)^2+D/(x-√2/2),解方程组即可得到A=3/4 B=√2/4 C=3/4 D=-√2/4
积分(x^2+1)/(x^4+1)dx=积分(3/4)/(x+√2/2)^2+(√2/4)/(x+√2/2)d(x+√2/2) +(3/4)/(x-√2/2)^2-(√2/4)/(x-√2/2)d(x-√2/2)=√2/4ln(x+√2/2)-√2/4ln(x-√2/2)-(3/4)/(x+√2/2)-(3/4)/(x+√2/2)+C
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