求解不定积分 下面怎么算 40
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原积分式 = [(x-3)ln(x-3)] * [ln(x-3)-2] + 2x + C’(C' 是常数)
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∫ x d[ln^2(x-3)]
= ∫ x [2ln(x-3)] [1/(x-3)] dx
= ∫ [1-3/(x-3)] [2ln(x-3)] dx
= ∫ 2ln(x-3) + [6ln(x-3)]/(x-3) dx
= [ 2(x-3)ln(x-3)-2x ] + ∫ 6ln(x-3) d[ln(x-3)]
= 2(x-3)ln(x-3)-2x + 3ln^2(x-3) + C..............(C 是常数)
因此原积分式
= xln^2(x-3) - ∫ x d[ln^2(x-3)]
= xln^2(x-3) - 2(x-3)ln(x-3) + 2x - 3ln^2(x-3) + C'
= [(x-3)ln(x-3)] * [ln(x-3)-2] + 2x + C'............... (C' = -C)
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∫ x d[ln^2(x-3)]
= ∫ x [2ln(x-3)] [1/(x-3)] dx
= ∫ [1-3/(x-3)] [2ln(x-3)] dx
= ∫ 2ln(x-3) + [6ln(x-3)]/(x-3) dx
= [ 2(x-3)ln(x-3)-2x ] + ∫ 6ln(x-3) d[ln(x-3)]
= 2(x-3)ln(x-3)-2x + 3ln^2(x-3) + C..............(C 是常数)
因此原积分式
= xln^2(x-3) - ∫ x d[ln^2(x-3)]
= xln^2(x-3) - 2(x-3)ln(x-3) + 2x - 3ln^2(x-3) + C'
= [(x-3)ln(x-3)] * [ln(x-3)-2] + 2x + C'............... (C' = -C)
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