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计算:定积分∫(在上2,在下 0)ln(x+√x^2+1) dx 注 跟号包括x^2+1 ..
人气:277 ℃ 时间:2020-03-29 05:23:16
解答
∫[0,2] ln[x+√(x^2+1)] dx
=xln[x+√(x^2+1)][0,2]-∫[0,2] xdln[x+√(x^2+1)]
=2ln(2+√5)-∫[0,2] x/[x+√(x^2+1)]*[x+√(x^2+1)]'dx
=2ln(2+√5)-∫[0,2] x/[x+√(x^2+1)]*[1+x/√(x^2+1)]dx
=2ln(2+√5)-∫[0,2] x/√(x^2+1)dx
=2ln(2+√5)-√(x^2+1)[0,2]
=2ln(2+√5)-√5+1
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