What is #f(x) = int (3x)/(x-4) # if #f(2)=1 #?

Integrating this is a little tricky:

First, bring out the constant #3#:

#f(x)=3intx/(x-4)dx#

From here, either long divide #x# and #x-4# or split up the terms as follows for the same result:

#=3int(x-4+4)/(x-4)dx=3int(x-4)/(x-4)+4/(x-4)dx#

#=3int1+4/(x-4)dx#

Split up the integrals:

#=3int1dx+3int4/(x-4)dx#

Here, remember that the #3# should be multiplied into both integral, not just the first. We can easily evaluate the first integral. As an aside, we don't need to show the constant of integration until we evaluate the other integral as well.

#=3x+3int4/(x-4)dx#

Bring out the #4#:

#=3x+12int1/(x-4)dx#

Notice that we have an integral in the form #int(du)/u#, since the derivative of the denominator is the fraction's numerator (the derivative of #x-4# is #1#).

Since #int(du)/u=lnabsu+C#, we get

#f(x)=3x+12lnabs(x-4)+C#

Now, we can use the original condition #f(2)=1# to solve for #C#.

#1=3(2)+12lnabs(2-4)+C#

#1=6+12lnabs(-2)+C#

#-5=12ln2+C#

#-5-12ln2=C#

Plug the value of #C# into the what we found #f(x)# to be:

#f(x)=3x+12lnabs(x-4)-5-12ln2#