求解2道不定积分的题目.DX/√(X+1)+√(X-1)DX/1+E^X
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求解2道不定积分的题目.DX/√(X+1)+√(X-1)DX/1+E^X
求解2道不定积分的题目.
DX/√(X+1)+√(X-1)
DX/1+E^X
求解2道不定积分的题目.DX/√(X+1)+√(X-1)DX/1+E^X
1/[√(x + 1) + √(x - 1)] * [√(x + 1) - √(x - 1)]/[√(x + 1) - √(x - 1)],分母有理化
= [√(x + 1) - √(x - 1)]/[(x + 1) - (x - 1)]
= (1/2)[√(x + 1) - √(x - 1)]
∫ dx/[√(x + 1) + √(x - 1)] = (1/2)∫ √(x + 1) dx - (1/2)∫ √(x - 1) dx
= (1/2) (2/3)(x + 1)^(3/2) - (1/2) (2/3)(x - 1)^(3/2) + C
= (1/3)(x + 1)^(3/2) - (1/3)(x - 1)^(3/2) + C
∫ dx/(1 + e^x)
= ∫ (1 + e^x - e^x)/(1 + e^x) dx
= ∫ dx - ∫ e^x/(1 + e^x) dx
= x - ∫ d(e^x + 1)/(1 + e^x)
= x - ln(1 + e^x) + C
∫ dx/[√(x+1)+√(x-1)]
= ∫ [√(x+1)-√(x-1)]dx / [√(x+1)+√(x-1)][√(x+1)-√(x-1)] 分母有理化
= ∫ ½[√(x+1)-√(x-1)]dx
= ½ 2/3[(x+1)^(3/2)-(x-1)^(3/2)] + C
= 1/3[(x+1)^(3/2)-(x-1)^(3/2)]...
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∫ dx/[√(x+1)+√(x-1)]
= ∫ [√(x+1)-√(x-1)]dx / [√(x+1)+√(x-1)][√(x+1)-√(x-1)] 分母有理化
= ∫ ½[√(x+1)-√(x-1)]dx
= ½ 2/3[(x+1)^(3/2)-(x-1)^(3/2)] + C
= 1/3[(x+1)^(3/2)-(x-1)^(3/2)] + C
∫ dx/(1+e^x)
= ∫ (1 + e^x - e^x)dx / (1+e^x)
= ∫ (1+ e^x)/(1+e^x) dx - ∫ e^x / (1+e^x) dx
= ∫ dx - ∫ 1 / (1+e^x) d(1+e^x)
= x - ln(1+e^x) +C
收起
你们怎么打出不定积分的符号的?
输入法里不是有嘛,数学符号