Integrate (logx)/(x(1+logx)(2+logx)) ?

1 Answer
Teddy
Jun 20, 2018

#intlogx/(x(1+logx)(2+logx))dx=-log|logx+1|+2log|logx+2|+C#

Explanation:

#I=intlogx/(x(1+logx)(2+logx))dx#
Let #u=logx# so #du=1/xdx#
#I=intu/((u+1)(u+2))du#
Let's examine #u/((u+1)(u+2))# and its partial fraction decomposition:
#u/((u+1)(u+2))=A/(u+1)+B/(u+2)#
#u=A(u+2)+B(u+1)#
#u=-2->B=2#
#u=-1->A=-1#
Therefore, #u/((u+1)(u+2))=-1/(u+1)+2/(u+2)#
#I=int(-1/(u+1)+2/(u+2))du#
#I=-int1/(u+1)du+2int1/(u+2)du#
#I=-log|u+1|+2log|u+2|+C#
#I=-log|logx+1|+2log|logx+2|+C#

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