第1个回答 2011-11-27
∫(x*e^x)/√(e^x+1) dx
Let ψ=√(e^x+1) => x=ln(ψ²-1) => dx=2ψ/(ψ²-1) dψ
= ∫[ln(ψ²-1)*(ψ²-1)/ψ] * [2ψ/(ψ²-1)] dψ
= 2∫ln(ψ²-1) dψ
= 2ψln(ψ²-1) - 2∫ψ dln(ψ²-1)
= 2√(e^x+1)ln(e^x+1-1) - 2∫ψ*2ψ/(ψ²-1) dψ
= 2√(e^x+1)ln(e^x) - 4∫(ψ²-1+1)/(ψ²-1) dψ
= 2x√(e^x+1) - 4∫ dψ + 4∫dψ/(ψ²-1)
= 2x√(e^x+1) - 4ψ - 4(1/2)ln|(ψ-1)/(ψ+1)| + C
= 2x√(e^x+1) - 4√(e^x+1) - 2ln|[√(e^x+1)-1] / [√(e^x+1)+1]| + C
= 2(x-2)√(e^x+1) - 2ln|[√(e^x+1)-1] / [√(e^x+1)+1]| + C本回答被提问者采纳
Let ψ=√(e^x+1) => x=ln(ψ²-1) => dx=2ψ/(ψ²-1) dψ
= ∫[ln(ψ²-1)*(ψ²-1)/ψ] * [2ψ/(ψ²-1)] dψ
= 2∫ln(ψ²-1) dψ
= 2ψln(ψ²-1) - 2∫ψ dln(ψ²-1)
= 2√(e^x+1)ln(e^x+1-1) - 2∫ψ*2ψ/(ψ²-1) dψ
= 2√(e^x+1)ln(e^x) - 4∫(ψ²-1+1)/(ψ²-1) dψ
= 2x√(e^x+1) - 4∫ dψ + 4∫dψ/(ψ²-1)
= 2x√(e^x+1) - 4ψ - 4(1/2)ln|(ψ-1)/(ψ+1)| + C
= 2x√(e^x+1) - 4√(e^x+1) - 2ln|[√(e^x+1)-1] / [√(e^x+1)+1]| + C
= 2(x-2)√(e^x+1) - 2ln|[√(e^x+1)-1] / [√(e^x+1)+1]| + C本回答被提问者采纳
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