Use a direct substitution by let u(x) = 7 + sin(5x). Then du = 5cos(5x) dx. Then (1/5) du = cos(5x) dx. Then make the replacements.

dy/dx = (cos (5x)) / (7 + sin (5x))

y = INT {cos (5x) / (7 + sin (5x))} dx * I don't know how to make the integral sign on here *

y = INT {(1/5) 1/u} du

y = (1/5) ln |u| + C

y = (1/5) ln |7 + sin(5x)| + C

To find C, let y = 6 and x = 0.

6 = (1/5) ln |7 + sin(5*0)| + C

6 = (1/5) ln |7 + 0| + C

6 = (1/5) ln 7 + C

6 - 1/5 ln 7 = C

Then y = (1/5) ln |7 + sin(5x)| + 6 - 1/5 ln 7

y = (1/5) ln |(7 + sin(5x))/7| + 6

y = (1/5) ln |1 + (1/7)sin(5x)| + 6

As for the second question, simply let y = f(x) and so then dy/dx = f ' (x). Then you see that INT {f'(x) / f(x)} dx = INT {1/y} dy = ln |y| + C = ln |f(x)| + C. Since f(x) > 0 we see that |f(x)| = f(x). Thus, we see the answer is ln (f(x)) + C as needed.