What is #f(x) = int x/(x-1) dx# if #f(2) = 0 #?

1 Answer
Costas C.
May 26, 2016

Since #ln# can't help you, set the denominator because of its simple form as a variable. When you solve the integral, just set #x=2# to fit the #f(2)# in the equation and find the integration constant.

Answer is:

#f(x)=x+ln|x-1|-2#

Explanation:

#f(x)=intx/(x-1)dx#

The #ln# function will not help in this case. However, since the denominator is quite simple (1st grade):

Set #u=x-1=>x=u+1#

and #(du)/dx=d(x+1)/dx=(x+1)'=1=>(du)/dx=1<=>du=dx#

#intx/(x-1)dx=int(u+1)/(u)du=int(u/u+1/u)du=#

#=int(1+1/u)du=int1du+int(du)/u=u+ln|u|+c#

Substituting #x# back:

#u+ln|u|+c=x-1+ln|x-1|+c#

So:

#f(x)=intx/(x-1)dx=x-1+ln|x-1|+c#

#f(x)=x-1+ln|x-1|+c#

To find #c# we set #x=2#

#f(2)=2-1+ln|2-1|+c#

#0=1+ln1+c#

#c=-1#

Finally:

#f(x)=x-1+ln|x-1|+c=x-1+ln|x-1|-1=x+ln|x-1|-2#

#f(x)=x+ln|x-1|-2#

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