Question 1

Please show me how to do this.

Question 2

Answer:

$f(x) = (1)/(2)(x - \ln (x - 1) - 2)$

Explanation:

f'(x) = (1)/(2) - (1)/(2x - 2)

$f(x) = \int ((1)/(2) - (1)/(2x - 2)) dx$

$f(x) = \int [(1)/(2) - (1)/(2(x - 2))] dx$

$f(x) = \int (1)/(2)dx - (1)/(2) \int (1)/(x - 1) dx$

$f(x) = (1)/(2)x - (1)/(2) \ln (x - 1) + C$

$f(x) = (x - \ln (x - 1))/(2) + C$

$f(2) = (2 - \ln (2 - 1))/(2) + C = 0$

$f(2) = (2 - \ln (1))/(2) + C = 0$

f(2) = (2)/(2) + C = 0

f(2) = 1 + C = 0

C = -1

$f(x) = (x - \ln (x - 1))/(2) - 1$

$f(x) = (1)/(2)[x - \ln (x - 1) - 2]$

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