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

1 Answer
Noah G
Aug 10, 2017

f(x) = ln|((x - 3)/(x - 2))| + 6 - ln(4/3)f(x)=ln∣∣∣(x−3x−2)∣∣∣+6−ln(43)

Explanation:

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

f(x) = ln|x - 3| - ln|x - 2| + Cf(x)=ln|x−3|−ln|x−2|+C

f(x) = ln|(x- 3)/(x - 2)| + Cf(x)=ln∣∣∣x−3x−2∣∣∣+C

Now we determine the value of the constant of integration.

6 = ln|(-1 - 3)/(-1 - 2)| + C6=ln∣∣∣−1−3−1−2∣∣∣+C

6 = ln|4/3| + C6=ln∣∣∣43∣∣∣+C

C = 6 - ln(4/3)C=6−ln(43)

So the function is

f(x) = ln|((x - 3)/(x - 2))| + 6 - ln(4/3)f(x)=ln∣∣∣(x−3x−2)∣∣∣+6−ln(43)

Hopefully this helps!

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