ln x = Integral from 1 to x 1 / t dt where x must be a positive number. Suppose a definite integral has a variable for its upper limit of integration, for instance

= integral

(1,x) (sin t) dt
Evaluation the integral gives
-cos t with the limits of 1 to x = -cos x + cos 1 = -cos x + 0.5403....

The answer is an expression involving the upper limit of integration, x. Thus the integral is a function of x. The integral equals the area of the region under the graph of y = sin t from t = 1 to t = x. Clearly the area is a function of the value picked for x.