This should be a refresher from Calculus I, but I'll review what you should know about integrals at this point.

There are, broadly speaking, two big questions in calculus:

- How can we describe the way that something is changing?
- How can we describe the accumulation of a changing quantity?

The second question is answered with the integral, the main topic of much of Calculus II. Graphically, this looks like the area beneath the graph of a function \(f\):

The Fundamental Theorem of Calculus ties these two together, connecting a function \(f(x)\) and its antiderivative \(F(x)\) \(\big(\)in other words, \(F'(x) = f(x)\big)\): \[\text{If } F(x) = \int_a^x f(t) \ dt \text{, then } F'(x) = f(x) \text{ and } \int_a^b f(x) \ dx = F(b) - F(a)\]

This is a profound result, tying together the geometric idea of the integral as the area under the curve to the analytical idea of the integral as the antiderivative. This is how we begin to evaluate integrals, by working backwards from known derivatives.

In Calculus I, you learned how to differentiate functions like \(x^3\): pull the power down in front, and subtract one from the exponent. \[\dfrac{d}{dx} \left[x^3\right] = 3x^2\] Therefore, to take the antiderivative of a power function, reverse the process by adding one to the exponent and dividing by that new exponent: \[\int x^4 \ dx = \dfrac{1}{5}x^5 +C.\] In general, this "reverse Power Rule" looks like this: \[\ans{\int x^n \ dx = \dfrac{1}{n+1}x^{n+1} + C.}\] Notice that we included \(+C\) at the end of this answer. In case you've forgotten, that's because, for instance, taking the derivative of \[\dfrac{1}{5}x^5 + C\] will be \(x^4\) for any constant \(C\), so whenever we antidifferentiate, we account for this by adding the arbitrary constant \(C\).

This reverse power rule works for any power function (including functions that may not initially look like power functions, like \(\sqrt{x} = x^{1/2}\) or \(1/x^3 = x^{-3}\)), with one notable exception: the function \(f(x)=1/x\). Why is this? Try to evaluate this integral using the rule and see what happens: \[\int \dfrac{1}{x} \ dx = \int x^{-1} \ dx = \dfrac{1}{0}x^0 + C\] Since this answer includes division by 0, it is not valid. Therefore, we have to handle this case separately. However, this reverse power rule gives us the ability to integrate a broad class of functions (pretty easily, I might add).

Okay, so let's take care of this reciprocal function. To do so, we simply have to remember that \[\dfrac{d}{dx} \big[\ln x\big] = \dfrac{1}{x},\] so when we antidifferentiate the reciprocal function, we get the natural log function: \[\ans{\int \dfrac{1}{x} \ dx = \ln x + C.}\]

This is the easiest function to differentiate, and thus the easiest function to integrate: \[\dfrac{d}{dx} [e^x] = e^x \text{, so } \ans{\int e^x \ dx = e^x + C.}\]

In Calc I, you just had to bite the bullet and memorize these derivatives, and now you can take advantage of that to reverse the process: \[\begin{align} \dfrac{d}{dx} (\sin x) = \cos x &\longrightarrow \ans{\int \cos x \ dx = \sin x + C}\\ \dfrac{d}{dx} (\cos x) = -\sin x &\longrightarrow \ans{\int \sin x \ dx = -\cos x + C}\\ \dfrac{d}{dx} (\tan x) = \sec^2 x &\longrightarrow \int \sec^2 x \ dx = \tan x + C\\ \dfrac{d}{dx} (\sec x) = \sec x \ \tan x &\longrightarrow \int \sec x \ \tan x \ dx = \sec x + C\\ \dfrac{d}{dx} (\csc x) = -\csc x \ \cot x &\longrightarrow \int -\csc x \ \cot x \ dx = \csc x + C\\ \dfrac{d}{dx} (\cot x) = -\csc^2 x &\longrightarrow \int -\csc^2 x \ dx = \cot x + C\\ \end{align}\] I boxed in the ones that you should definitely memorize; the others may reappear later, but most people don't bother memorizing them, instead looking them up if necessary.

Finally, remember a few basic rules: specifically, dealing with constants and addition.

Just like with derivatives, constants can be pulled out of an integral, so that what remains can be integrated. In essence, constant multiples just get carried along for the ride, not changing the integral. \[\ans{\int cf(x) \ dx = c\int f(x) \ dx}\] \[\text{ex: } \int 3\sin x \ dx = 3\int \sin x \ dx = -3\cos x + C\] (Technically, that answer should be \(-3\cos x + 3C\), but since \(C\) is just an arbitrary constant, \(3C\) is also an arbitrary constant, so we can be a bit lazy with that.)

Again, just as with derivatives, we can split up a function into its terms (pieces that are added/subtracted together) and handle each term separately.

\[\ans{\int f(x) + g(x) \ dx = \int f(x) \ dx + \int g(x) \ dx}\] \[\text{ex: } \int x^2+2x \ dx = \int x^2 \ dx + \int 2x \ dx = \dfrac{1}{3}x^3 + x^2 + C\]