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1. I = (1/2)∫ ln(1+x^4)d(x^2) 令 x^2 = u
= (1/2)∫ ln(1+u^2)du
= (1/2)uln(1+u^2) - ∫ u^2du/(1+u^2)
= (1/2)uln(1+u^2) - ∫ [1-1/(1+u^2)]du
= (1/2)uln(1+u^2) - u + arctanu + C
= (1/2)x^2ln(1+x^4) - x^2 + arctan(x^2) + C
2. I = (1/3)∫arccosxd(x^3)
= (1/3)x^3arccosx + (1/3)∫x^3dx/√(1-x^2),
其中 J = (1/3)∫x^3dx/√(1-x^2) = (1/6)∫x^2d(x^2)/√(1-x^2)
令 √(1-x^2) = u, 则 1-x^2 = u^2, x^2 = 1-u^2,
J = (1/6)∫(1-u^2)(-2udu)/u = (-1/3)∫(1-u^2)du
= -u/3 + (1/9)u^3 + C = -√(1-x^2)/3 + (1/9)(1-x^2)^(3/2) + C
I = (1/3)x^3arccosx - √(1-x^2)/3 + (1/9)(1-x^2)^(3/2) + C
= (1/2)∫ ln(1+u^2)du
= (1/2)uln(1+u^2) - ∫ u^2du/(1+u^2)
= (1/2)uln(1+u^2) - ∫ [1-1/(1+u^2)]du
= (1/2)uln(1+u^2) - u + arctanu + C
= (1/2)x^2ln(1+x^4) - x^2 + arctan(x^2) + C
2. I = (1/3)∫arccosxd(x^3)
= (1/3)x^3arccosx + (1/3)∫x^3dx/√(1-x^2),
其中 J = (1/3)∫x^3dx/√(1-x^2) = (1/6)∫x^2d(x^2)/√(1-x^2)
令 √(1-x^2) = u, 则 1-x^2 = u^2, x^2 = 1-u^2,
J = (1/6)∫(1-u^2)(-2udu)/u = (-1/3)∫(1-u^2)du
= -u/3 + (1/9)u^3 + C = -√(1-x^2)/3 + (1/9)(1-x^2)^(3/2) + C
I = (1/3)x^3arccosx - √(1-x^2)/3 + (1/9)(1-x^2)^(3/2) + C
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