General Antiderivative

1. Homework Statement

Hi everyone. My Calc 1 final exam is tomorrow and due to some weather related issues we were not able to cover all material for this quarter. With that said, my professor gave us a take home quiz on material that was unable to be covered. I have done by best, but I am getting hung up on an antiderivative problem.

Find the general antiderivative of:

f'(x) = 1-2x-4/$$\sqrt{x}$$+5/x-8/(1+x^2)+9/x^4

2. Homework Equations

Now due to having virtually no time to learn about antiderivatives (we lost a whole week due to a blizzard and instructor illness) I am really unsure where to go. Do I need to rewrite the problem on one line and then find the opposite of the derivative?

Thanks!

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malawi_glenn
Homework Helper
well yes, what function, if you differentiate it, will give that expression as its derivative?

So basically the definition of an antiderivative is the oppositive of the derivative?

Wait, I did not just type that last sentence. =)

Do you mean:

$$f'(x) = 1-2x-4\sqrt{x}+\frac{5}{x}-\frac{8}{1+x^2}+\frac{9}{x^4}$$ ?

Basically when you're asked to find the anti-derivative you're trying to find the function f(x), which has this derivative f'(x), which is given.

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1- 2x -4$$\sqrt{x}$$+$$\frac{5}{x}$$-$$\frac{8}{1+x^{2}}$$+$$\frac{9}{x^{4}}$$

malawi_glenn
Homework Helper
yes!

But you often write it given f(x), find its primitive function F(x): F' = f

That is sort of how I figured that it would work. My derivative skills are giving me the most trouble at this point.

Do you have a textbook? First, read the section on anti-derivatives.

Hint: separate each quantity
f '(x) = 1, then f(x) = x + C.
f '(x) = 2x, then f(x) = ?
f '(x) = 9x^-4, then f(x) = ?
.
.
.
and so on.

f '(x) = 2x, then f(x) = x^2
f '(x) = 9x^-4, then f(x) = 9x^-3/-3
correct?

The general antiderivative will have a + C at the end.

f '(x) = 2x, then f(x) = x^2
f '(x) = 9x^-4, then f(x) = 9x^-3/-3
correct?
So far, so good

But Snazzy is right you need to add a +C (constant of integration) term to the anti derivative, because if f(x) had a constant value some where in it, like f(x) = 2x+5, f'(x) = 2 -- so when we integrate f'(x) we need to account for the 5. We don't necessairly know it's a 5 so that's why we add the +C

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It's so easy to forget the C at the end.