What is #f(x) = int 1/(x-3)-1/(x-2) dx# if #f(-1)=6 #?
1 Answer
Aug 10, 2017
Explanation:
We use the commonly used result
#f(x) = ln|x - 3| - ln|x - 2| + C#
#f(x) = ln|(x- 3)/(x - 2)| + C#
Now we determine the value of the constant of integration.
#6 = ln|(-1 - 3)/(-1 - 2)| + C#
#6 = ln|4/3| + C#
#C = 6 - ln(4/3)#
So the function is
#f(x) = ln|((x - 3)/(x - 2))| + 6 - ln(4/3)#
Hopefully this helps!