t^4－4t^3＋12t^2－32t－18＝0

F'(x)=1/√[1+(x^2)^4]*(x^2)'-1/√[1+(x^3)^4]*(x^3)'
=2x/√[1+(x^2)^4]-3x^2/√[1+(x^3)^4]
dF(x)={2x/√[1+(x^2)^4]-3x^2/√[1+(x^3)^4]}dx

t^2=u
∫(0,1) t√(1+t^2+t^4)dt
=(1/2)∫(0,1) √(1+u+u^2)du
=(1/2)∫(0,1) √(u+1/2)^2+3/4)du (用积分表√(u^2+a^2)du
=(1/2){[(u+1/2)/2]√(1+u+u^2)+(3/8)ln(u+1/2+√(1+u+u^2))}|(0,1)
代入上下限即可

哈哈,今天第三次做这个题了.你将下面的x换成t就行了.
∫ x²/(1+x^4) dx
=(1/2)∫ (x²-1+x²+1)/(1+x^4) dx
=(1/2)∫ (x²-1)/(1+x^4) dx + (1/2)∫ (x²+1)/(1+x^4) dx
分子分同除以x²
=(1/2)∫ (1-1/x²)/(1/x²+x²) dx + (1/2)∫ (1+1/x²)/(1/x²+x²) dx
分子放到微分之后
=(1/2)∫ 1/(1/x²+x²) d(x+1/x) + (1/2)∫ 1/(1/x²+x²) d(x-1/x)
=(1/2)∫ 1/(1/x²+x²+2-2) d(x+1/x) + (1/2)∫ 1/(1/x²+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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