What is the integral of #int (x-2)/(x-1)# from 0 to 2?

1 Answer
Mar 26, 2016

#2#

Explanation:

We can rewrite #x-2# as #x-1-1#. We do this so that it mimics the base.

#int_0^2(x-2)/(x-1)dx=int_0^2(x-1-1)/(x-1)dx#

#=int_0^2((x-1)/(x-1)-1/(x-1))dx=int_0^2(1-1/(x-1))dx#

All I've shown here is that #(x-2)/(x-1)=1-1/(x-1)#. You could also show this through polynomial long division or synthetic division.

To integrate this, I will split it into two indefinite integrals (for now). Later, we will come back and evaluate the combined integral from #0# to #2#.

We have:

#int1dx=x+C#

And, slightly trickier:

#-int1/(x-1)=-lnabs(x-1)+C#

For the previous integral, notice that the derivative of the denominator is present in the numerator. This means that we have a natural logarithm integral present.

Combining these, we want to evaluate

#=[x-lnabs(x-1)color(white)(""/"")]_0^2#

#=(2-lnabs(1))-(0-lnabs(-1))#

#=(2-0)-(0-0)=2#