∫ln(x+a)/(x+b)dx

网友回答

答：∫In(x+a)/(x+b)dx
=∫[(In(x+a)-In(x+b)]dx
=∫In(x+a)dx-∫In(x+b)dx
=(x+a)[In(x+a)-1]-(x+b)[In(x+b)-1]+C
=(x+a)In(x+a)-(x+b)In(x+b)+b-a+C
C是常数.这里说一下,
∫In(x+a)dx=∫In(x+a)(x)'dx
=xIn(x+a)-∫{[In(x+a)]'*x}dx
=xIn(x+a)-∫[x/(x+a)]dx
=xIn(x+a)-∫[1-a/(x+a)]dx
=xIn(x+a)-x+∫a/(x+a)dx
=xIn(x+a)-x+aIn(x+a)+C
=(x+a)[In(x+a)-1]+a+C
a,C看成常数,
(x+a)[In(x+a)-1]+a+C与(x+a)[In(x+a)-1}+C是等价的.
这是定积分分部积分法.