计算不定积分,{x+2/（根号x的平方－2x－1） 各位大侠、、真的是急用啊 、、、
2个回答
∫(x+2)dx/√(x^2-2x-1)
=∫(x-1)dx/√(x^2-2x-1) +∫3dx/√[(x-1)^2-2]
=(1/2)∫d(x^2-2x-1)/√(x^2-2x-1)+(6√2)∫d(x/√2-1/√2)/√[(x/√2-1/√2)^2-1]
--取sect=(x/√2-1/√2)
(sint)^2=1-(1/sect)^2=1-1/[2(x-1)^2]=[2x^2-4x+3]/[2(x-1)^2]
sint=√(x^2-2x+3/2)/(x-1)^2
=√(x^2-2x-1)+6√2 ∫dsect/√[(sect)^2-1]
=√(x^2-2x-1)+6√2∫dsint/[(1-sint)(1+sint)]
=√(x^2-2x-1)+3√2ln[(1+sint)/(1-sint)]+C
=√(x^2-2x+1)+3√2ln [(x-1)^2+√(x^2-2x+3/2)]/[(x-1)^2-√(x^2-2x+3/2)] +C
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