微积分计算题!
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(1) Integrate[1/(x Ln[x]), x] = Integrate[1/Ln[x], Ln[x]] = Ln[Ln[x]]+c
(2) Integrage[E^t/(E^2t+1),t]=Ingegrate[1/(E^2t+1),E^t]=arctan[E^t]
Integrate[Sqrt[Cos[x]-Cos[x]^3],{x,-Pi/2,Pi/2}]=Integrate[Sqrt[Cos[x] Sin[x]^2],{x,-Pi/2,Pi/2}]
=-Integrate[Sin[x] Sqrt[Cos[x]],{x,-Pi/2,0}]+Integrate[Sin[x] Sqrt[Cos[x]],{x,0,Pi/2}]
=Integrage[Sqrt[Cos[x]],{Cos[x],-Pi/2,0}]-Integrate[Sqrt[Cos[x]],{Cos[x],0,Pi/2}]
=2/3 Cos[x]^(3/2)(此处把积分限代入)=4/3
lim[Integrate[(x-t)f[t],t]/x Integrate[f[x-t],t]]=lim[(x Integrate[f[t],t]-Integrate[t f[t],t])/xIntegrate[f[t],t]]
=lim[Integrate[f[t],t]/(Integrate[f[t],t]+x f[x])]
=f[0] x/(f[0] x+x f[0]) (此处用了连续函数的性质)
=1/2
令z=y^(-1)
z'=-y^-2 y'
y'=-z'/z^2
z'=-2 x/(x^2-1) 代入原方程
z=Integrate[-2 x/(x^2-1),x]=-Ln[Abs[x^2-1]]+c
y=1/z=1/(-Ln[Abs[x^2-1]]+c)
y(0)=1/c=1 c=1
y=1/(-Ln[Abs[x^2-1]]+1)
(2) Integrage[E^t/(E^2t+1),t]=Ingegrate[1/(E^2t+1),E^t]=arctan[E^t]
Integrate[Sqrt[Cos[x]-Cos[x]^3],{x,-Pi/2,Pi/2}]=Integrate[Sqrt[Cos[x] Sin[x]^2],{x,-Pi/2,Pi/2}]
=-Integrate[Sin[x] Sqrt[Cos[x]],{x,-Pi/2,0}]+Integrate[Sin[x] Sqrt[Cos[x]],{x,0,Pi/2}]
=Integrage[Sqrt[Cos[x]],{Cos[x],-Pi/2,0}]-Integrate[Sqrt[Cos[x]],{Cos[x],0,Pi/2}]
=2/3 Cos[x]^(3/2)(此处把积分限代入)=4/3
lim[Integrate[(x-t)f[t],t]/x Integrate[f[x-t],t]]=lim[(x Integrate[f[t],t]-Integrate[t f[t],t])/xIntegrate[f[t],t]]
=lim[Integrate[f[t],t]/(Integrate[f[t],t]+x f[x])]
=f[0] x/(f[0] x+x f[0]) (此处用了连续函数的性质)
=1/2
令z=y^(-1)
z'=-y^-2 y'
y'=-z'/z^2
z'=-2 x/(x^2-1) 代入原方程
z=Integrate[-2 x/(x^2-1),x]=-Ln[Abs[x^2-1]]+c
y=1/z=1/(-Ln[Abs[x^2-1]]+c)
y(0)=1/c=1 c=1
y=1/(-Ln[Abs[x^2-1]]+1)
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一
∫﹙1/xlnx﹚dx=∫﹙1/lnx﹚dlnx=ln|lnx|+C
∫【e^t/[e^﹙2t﹚+1]×dt
=∫【1/[e^﹙2t﹚+1]×d﹙e^t﹚
=arctan﹙e^t﹚+C
二
∫〈-π/2→π/2〉√﹙cosx-cos³x﹚×dx
=∫〈-π/2→π/2〉√[cosx﹙1-cos²x﹚]×dx
=∫〈-π/2→π/2〉√﹙cosx×sin²x﹚×dx
=∫〈-π/2→0〉﹙-sinx﹚√﹙cosx﹚×dx+∫〈0→π/2〉sinx√﹙cosx﹚×dx
=∫〈-π/2→0〉√﹙cosx﹚×d﹙cosx﹚-∫〈0→π/2〉√﹙cosx﹚×d﹙cosx﹚
=2/3×﹙cosx﹚^﹙3/2﹚|〈-π/2→0〉-2/3×﹙cosx﹚^﹙3/2﹚|〈0→π/2〉
=4/3
三
lim〈0→x〉【∫〈0→x〉﹙x-t﹚f﹙t﹚dt】/【x∫〈0→x〉f﹙x-t﹚dt】这是0/0型用罗比达法则
=lim〈0→x〉【x∫〈0→x〉f﹙t﹚dt- ∫〈0→x〉tf﹙t﹚dt】/【x∫〈0→x〉f﹙x-t﹚dt】
=lim〈0→x〉【∫〈0→x〉f﹙t﹚dt+xf﹙x﹚-xf﹙x﹚】/【∫〈0→x〉f﹙x-t﹚dt+xf﹙0﹚】
=lim〈0→x〉【∫〈0→x〉f﹙t﹚dt】/【∫〈0→x〉f﹙x-t﹚dt+xf﹙0﹚】仍是0/0型
=lim〈0→x〉f﹙x﹚/【f﹙0﹚+f﹙0﹚+0】
=1/2
四
﹙x²-1﹚y'+2xy²=0
∴﹙x²-1﹚dy/dx=-2xy²
∴dy/y²=-2x/﹙x²-1﹚dx=2x/﹙1-x²﹚dx
∴∫dy/y²=∫[2x/﹙1-x²﹚]dx
∴-1/y=∫[1/﹙1-x²﹚]dx²=-∫[1/﹙1-x²﹚]d﹙1-x²﹚=-ln|1-x²|+C
∴1/y=ln|1-x²|+C
∫﹙1/xlnx﹚dx=∫﹙1/lnx﹚dlnx=ln|lnx|+C
∫【e^t/[e^﹙2t﹚+1]×dt
=∫【1/[e^﹙2t﹚+1]×d﹙e^t﹚
=arctan﹙e^t﹚+C
二
∫〈-π/2→π/2〉√﹙cosx-cos³x﹚×dx
=∫〈-π/2→π/2〉√[cosx﹙1-cos²x﹚]×dx
=∫〈-π/2→π/2〉√﹙cosx×sin²x﹚×dx
=∫〈-π/2→0〉﹙-sinx﹚√﹙cosx﹚×dx+∫〈0→π/2〉sinx√﹙cosx﹚×dx
=∫〈-π/2→0〉√﹙cosx﹚×d﹙cosx﹚-∫〈0→π/2〉√﹙cosx﹚×d﹙cosx﹚
=2/3×﹙cosx﹚^﹙3/2﹚|〈-π/2→0〉-2/3×﹙cosx﹚^﹙3/2﹚|〈0→π/2〉
=4/3
三
lim〈0→x〉【∫〈0→x〉﹙x-t﹚f﹙t﹚dt】/【x∫〈0→x〉f﹙x-t﹚dt】这是0/0型用罗比达法则
=lim〈0→x〉【x∫〈0→x〉f﹙t﹚dt- ∫〈0→x〉tf﹙t﹚dt】/【x∫〈0→x〉f﹙x-t﹚dt】
=lim〈0→x〉【∫〈0→x〉f﹙t﹚dt+xf﹙x﹚-xf﹙x﹚】/【∫〈0→x〉f﹙x-t﹚dt+xf﹙0﹚】
=lim〈0→x〉【∫〈0→x〉f﹙t﹚dt】/【∫〈0→x〉f﹙x-t﹚dt+xf﹙0﹚】仍是0/0型
=lim〈0→x〉f﹙x﹚/【f﹙0﹚+f﹙0﹚+0】
=1/2
四
﹙x²-1﹚y'+2xy²=0
∴﹙x²-1﹚dy/dx=-2xy²
∴dy/y²=-2x/﹙x²-1﹚dx=2x/﹙1-x²﹚dx
∴∫dy/y²=∫[2x/﹙1-x²﹚]dx
∴-1/y=∫[1/﹙1-x²﹚]dx²=-∫[1/﹙1-x²﹚]d﹙1-x²﹚=-ln|1-x²|+C
∴1/y=ln|1-x²|+C
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