Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 68

You need to evaluate the function f using the provided information, hence, you need to apply the antiderivative, such that:
int (f'(x) sinhx + 2cosh x)dx = int (f'(x)(e^x - e^(-x))/2) + (e^x + e^(-x)))dx
int (f'(x) sinhx + 2cosh x)dx = (1/2)int (f'(x)(e^x)dx - (1/2)int f'(x) e^(-x)dx + int e^x dx + int e^(-x)dx
You need to evaluate each of the two integrals separately. You may integrate int (f'(x)(e^x)dx using integration by parts. You may consider u = e^x and dv = f'(x), such that:
int udv = uv - int vdu
u = e^x => du = e^x
dv = f'(x) => v = f(x)
int f'(x) e^x dx = f(x) e^x - int f(x)e^x => (1/2)int f'(x) e^x dx = (1/2)f(x) e^x - (1/2)int f(x)e^x
(1/2)int f'(x) e^(-x)dx = (1/2)f(x)e^(-x) + (1/2)int e(-x)f(x) dx
u = e^(-x) => du = -e^(-x)
dv = f'(x) => v = f(x)
(1/2)int (f'(x)(e^x)dx - (1/2)int f'(x) e^(-x)dx= (1/2)f(x) e^x - (1/2)int f(x)e^x - (1/2)f(x)e^(-x) - (1/2)int e(-x)f(x) dx
int e^x dx + int e^(-x)dx = e^x - e^(-x) + c
Hence, evaluating the integral, yields:
int (f'(x) sinhx + 2cosh x)dx = (1/2)f(x) e^x - (1/2)int f(x)e^x - (1/2)f(x)e^(-x) - (1/2)int e^(-x)f(x) dx + e^x - e^(-x) + c
int (f'(x) sinhx + 2cosh x)dx = (1/2)f(x)(e^x - e^(-x)) - (1/2)int (f(x)e^x -e(-x)f(x))dx + e^x - e^(-x) + c
int (f'(x) sinhx + 2cosh x)dx = f(x)sinh x - int f(x) sinh x + 2sinhx + c
Given the provided information, this is the answer for the antiderivative int (f'(x) sinhx + 2cosh x)dx = f(x)sinh x - int f(x) sinh x + 2sinhx + c.