Question

求不定积分∫(x^2/(1+x^4))dx

Answer 1

这是个技巧比较高的积分,不是楼上答得那么简单的
∫ x²/(1+x⁴) dx
=(1/2)∫ (x²-1+x²+1)/(1+x⁴) dx
=(1/2)∫ (x²-1)/(1+x⁴) dx + (1/2)∫ (x²+1)/(1+x⁴) dx
分子分母同除以x²
=(1/2)∫ (1-1/x²)/(x²+1/x²) dx + (1/2)∫ (1+1/x²)/(x²+1/x²) dx
分子放到微分之后,然后分母凑个2出来
=(1/2)∫ 1/(x²+1/x²+2-2) d(x+1/x) + (1/2)∫ 1/(x²+1/x²-2+2) d(x-1/x)
=(1/2)∫ 1/[(x+1/x)²-2] d(x+1/x) + (1/2)∫ 1/[(x-1/x)²+2] d(x-1/x)
=(√2/8)ln|(x+1/x-√2)/(x+1/x+√2)| + (√2/4)arctan[(x-1/x)/√2] + C
=(√2/8)ln|(x²+1-√2x)/(x²+1+√2x)| + (√2/4)arctan[(x-1/x)/√2] + C

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