What is #f(x) = int 1/(x-3)-1/(x-2) dx# if #f(-1)=6 #?

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Answer 1 · 2017-08-10T02:08:49

#f(x) = ln|((x - 3)/(x - 2))| + 6 - ln(4/3)#

We use the commonly used result #int (1/x) dx = ln|x|# to solve.

#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!