How to antidifferentiate with a negative exponent? $6x^{-2} + 2x^{-4} -3x^{-3}$.
How would the anti diff process work for fractions? I have $\frac{1}{6x^2}$ for the first part. Am I supposed to add one to x, and then put it all over 3? How would the fraction then resolve?
 A: Hint. A general rule, working for all exponents (both negative and non-negative):

$$
f(x)=x^{\alpha} \quad \text{gives an antiderivative } \, F(x)=\frac{x^{\alpha+1}}{\alpha+1}+C \quad \text{if} \quad \alpha \neq-1,
$$
$$
f(x)=x^{-1}= \frac1x\quad \text{gives an antiderivative } \, F(x)=\ln (x)+C \quad \text{if} \quad x>0,
$$ 

where $C$ is any constant.
A: Two general formulas:    
$(x^n)' = n \cdot x^{n-1}$ (for any integer $n \neq 0$)
$(c \cdot f(x))' = c \cdot f'(x)$  (for any number $c$)     
So we know that: $(x^{-1})' = -1 \cdot x^{-2}$
So:  $(-6 \cdot x^{-1})' = -6 \cdot (-1) \cdot x^{-2} = 6 \cdot x^{-2}$
So the antiderivative of $6 \cdot x^{-2}$ is $-6 \cdot x^{-1}$.   
A: Just continue to use the rule $(x^{n})'=nx^{n-1}$ for all $n\in \mathbb{Z}$, $n\not=0$. Hence, 
$(x^{-1})'=-x^{-2}\implies(-6x^{-1})'=6x^{-2}$. 
$(x^{-3})'=-3x^{-4}\implies(-\frac{2}{3}x^{-3})'=2x^{-4}$. 
$(x^{-2})'=-2x^{-3}\implies(\frac{3}{2}x^{-2})'=-3x^{-3}$.
So, the antiderivative is: $-6x^{-1}-\frac{2}{3}x^{-3}+\frac{3}{2}x^{-2}$. 
A: I think your confusion comes from missing this:
$\frac{1}{x^2}=x^{-2}$
$\frac{1}{x^3}=x^{-3}$
$\frac{1}{x^4}=x^{-4}$
Just proceed as you would for positive powers, but being careful about minus signs.
$\frac{1}{x}=x^{-1} $ is a special case since you can't get it by differentiating $x^0$ (try it!) but they've very kindly avoided putting that in the expression.
Example:
Find the antiderivative of $\frac{2}{x^5}$.
First write this as $$\frac{2}{x^5}=2x^{-5}$$
Thinking of this as $2x^{n}$, what is $n$? $$n=-5$$
So, what is $n+1$? $$n+1=-4$$
In which case, what is $\frac{x^{n+1}}{n+1}$?
$$\frac{x^{n+1}}{n+1}=\frac{x^{-4}}{-4} = -\frac{1}{4{x^4}}$$
But we wanted the antiderivative of $2x^{-5}$, not just $x^{-5}$, so we still need to multiply by 2 and we finally get $$2\left(-\frac{1}{4{x^4}}\right)=-\frac{1}{2{x^4}}$$
