求 x4/(1+x4) 的不定积分

∫x^4/(1+x^4)dx=∫ {1-[1/(1+x^4)]}dx
又
∫dx/(1+x^4)
=(1/2)[∫(1+x²)dx/(1+x^4)+∫(1-x²)dx/(1+x^4)] . 分子分母同除于x²
=(1/2){∫[(1/x²)+1]dx/(1/x²+x²)+∫[(1/x²)-1]dx/(1/x²+x²)}
=(1/2){∫[d[x-(1/x)]/[(x-1/x)²+2]-∫d[x+(1/x)]/[(x+1/x)²-2]}
=(1/2){(1/√2)arctan[(x-1/x)/√2]-(1/2√2)ln|(x+1/x-√2)/(x+1/x+√2)|}+C
=[1/(2√2)]arctan[(x²-1)/x√2]-(1/4√2)ln[(x²-x√2+1)/(x²+x√2+1)]+C
所以∫x^4/(1+x^4)dx=∫ {1-[1/(1+x^4)]}dx
=x-[1/(2√2)]arctan[(x²-1)/x√2]+(1/4√2)ln[(x²-x√2+1)/(x²+x√2+1)]+C1
C1=-C均为任意常数
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