换元积分法求解:∫(√(4x-5))dx

2020-09-17 社会 67阅读
∫x/√(5-4x)dx
=-(1/4)∫(5-4x-5)/√(5-4x)dx
=-(1/4)∫[√(5-4x) - 5/√(5-4x)]dx
=(1/16)∫[√(5-4x) - 5/√(5-4x)]d(5-4x)
=(1/16)[(2/3)(5-4x)^(3/2) - 10√(5-4x)]
=(1/16)[(2/3)(1-27) - 10(1-3)]
=1/6
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