Question

∫dx/(1+ x^4)怎么积分？

Answer 1

∫dx/(1+x^4) =(1/2)[∫(1+x²)dx/(1+x^4)+∫(1-x²)dx/(1+x^4)] （两个积分都是分子分母同除以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。
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Answer 2

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