The general antiderivative of #f(x)# is #F(x)+C#, where #F# is a differentiable function. All that means is that if you differentiate the antiderivative, you get the original function - so to find the antiderivative, you reverse the process of finding a derivative.

Sound confusing? Easier done than said. What we're doing is only taking the indefinite integral of #f(x)# - in other words, #int x-7 dx#. The properties of integrals say that we can break it up in pieces in cases of addition and subtraction; thus,

#intx-7dx = intxdx-int7dx#.

Further using the properties of integrals,

#intx-7dx = intxdx-7intdx#

First, let's do #intxdx#. What we're asking ourselves is: what function, when you take its derivative, equals #x#? Well, #x^2/2#, of course! Using the power rule, we multiply the expression by the exponent and then reduce the exponent by one; doing that gives #2*(x^(2-1))/2 = x#. So, our first integral reduces to #x^2/2+C#.

Now, why the #C#? We put the #C# (which is just a constant - any old number, like #2#, #sqrt(5)#, and #pi#) because we're finding the general antiderivative. Thus, we don't know if there's another number hiding in our antiderivative - so we put the #C# there to make it general and cover our behinds.

Finally, we evaluate #7intdx#. This (#intdx#) is called a perfect integral because its result is plain ol' #x#. Since we have a #7# in front of it, our final result is #7x+C# (never forget the #C#!).

We can finally put our pieces together for the final answer:

#intx-7dx = intxdx-int7dx#

#intx-7dx = (x^2/2 + C) - (7x + C#)

# = x^2/2 + C - 7x - C# (distributing the negative sign)

You might think #C-C = 0#, but that's not quite right. Recall that #C# is *any* number - both of them. So one #C# can be #4# and the other can be #3#, in which case #C-C = 1# or #-1#. But then again, #1# and #-1# are constants, right? In fact, #C-C# will always be a constant, and since #C# represents a constant, we can just call #C-C# normal #C#. Take my word for it.

Thus, the final result is #x^2/2-7x+C#.