How do I find the integral #intx/(x-6)dx# ?

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Answer 1 · 2014-09-30T02:00:46

By Substitution,

#int x/{x-6}dx=x+6ln|x-6|+C#

Let us look at some details.

#int x/{x-6}dx#

by the sunstitution #u=x-6#,
#Rightarrow x=u+6#
#Rightarrow dx=du#

#=int {u+6}/u du#

by splitting the integrand,

#=int (1+6/u) du#

#=u+6ln|u|+C_1#

by putting #u=x-6# back in,

#=x-6+6ln|x-6|+C_1#

by letting #C=C_1-6#,

#=x+6ln|x-6|+C#

I hope that this was helpful.