1/x^2（1+x^2）dx的不定积分

Answer 1

如图，这道题拆成两项来做就可以了，然后分别求两个函数的不定积分，希望可以帮助你

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哇哇超棒啊谢谢你

哇哇超棒啊谢谢你

Answer 2

Answer 3

= ∫ d(x^2) / 2 / (1+ x^2)^1/2
= (1+ x^2)^1/2 + C

Answer 4

1+x-x^2 = 5/4-(x-1/2)^2
let
x-1/2 = (√5/2)sinu
dx =(√5/2)cosu du
∫x/√(1+x-x^2)dx
=-(1/2)∫(1-2x)/√(1+x-x^2)dx +(1/2)∫dx/√(1+x-x^2)
=-(1/2)√(1+x-x^2) +(1/2)∫dx/√(1+x-x^2)
=-(1/2)√(1+x-x^2) +(√5/5)∫ du
=-(1/2)√(1+x-x^2) +(√5/5)u + C
=-(1/2)√(1+x-x^2) +(√5/5)arcsin[(2x-1)/√5] +C

Answer 5

∫ 1/(x²+x+1)² dx
=∫ 1/[(x+1/2)²+3/4]² dx
令t=x+1/2,
dt=dx
=∫ 1/(t²+3/4)² dt
令t=√3/2*tan s,
dt=√3/2*sec²s ds
=√3/2*∫ sec²s/(3/4*tan²s+3/4)² ds
=8/(3√3)*∫ cos²s ds
=8/(3√3)*1/2*∫ (1+cos2s) ds
=4/(3√3)*∫ dp+4/(3√3)*∫ cos2s ds
=4/(3√3)*s+4/(3√3)*1/2*sin2s+C
=4/(3√3)*s+4/(3√3)*sinscoss+C
=4/(3√3)*arctan(2t/√3)+4/(3√3)*2t/√(4t²+3)*√3/√(4t²+3)+C
=4/(3√3)*arctan(2t/√3)+4/(3√3)*(2√3*t)/(4t²+3)+C
=4/(3√3)*arctan[(2/√3)*(x+1/2)]+4/(3√3)*2√3*(x+1/2)/[4(x+1/2)²+3]+C
=4/(3√3)*arctan[(2x+1)/√3]+4/(3√3)*(2√3*x+√3)/[4(x²+x+1)]+C
=4/(3√3)arctan[(2x+1)/√3]+(1/3)(2x+1)/(x²+x+1)+C

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不是这道题😂