x的4次方加1的倒數(shù)的不定積分是?

數(shù)學(xué)人氣：585 ℃時間：2020-06-02 05:54:01

優(yōu)質(zhì)解答

∫ 1/(x⁴+1) dx
= (1/2)∫ [(x²+1)-(x²-1)]/(x⁴+1) dx
= (1/2)∫ (x²+1)/(x⁴+1) dx - (1/2)∫ (x²-1)/(x⁴+1) dx
= (1/2)∫ (1+1/x²)/(x²+1/x²) dx - (1/2)∫ (1-1/x²)/(x²+1/x²) dx
= (1/2)∫ d(x-1/x)/[(x-1/x)²+2] - (1/2)∫ d(x+1/x)/[(x+1/x)²-2]
= [1/(2√2)]arctan[(x-1/x)√2] - [1/(4√2)]ln|[(x+1/x-√2)/(x+1/x+√2)| + C
= [1/(2√2)]arctan[x/√2-1/(√2*x)] - [1/(4√2)]ln|(x²-√2*x+1)/(x²+√2*x+1)| + C