Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 71

You need to notice that if the derivative is a linear function, then the primitive is a quadratic function, such that:
f(x) = ax^2 + bx + c
Differentiating f(x) yields:
f'(x) = 2ax + b
You need to set equal 2x and 2ax + b , such that:
2ax + b = 2x
Comparing the coefficients of x both sides yields:
2a = 2 => a = 1
Notice that the free term b must be 0.
You should notice that c is not determined, yet. You may find out c using the information that the function passes through the point (1,0), such that:
f(1) = 0 => f(1) = a*1^2 + b*1 + c
a + b + c = 0
Replacing 1 for a and 0 for b yields:
1 + 0 + c = 0 => c = -1
Hence, evaluating the function under the given conditions, yields f(x) = x^2 - 1.