Standard integrals: e, ln, sin, cos, sec² and tan⁻¹ with a linear inside

To integrate a function means to find a function whose derivative is the one you started with. The answer is called an antiderivative. For example, the derivative of $x^3$ is $3x^2$, so an antiderivative of $3x^2$ is $x^3$.

Why "an" antiderivative and not "the"? Because $x^3 + 5$ and $x^3 - 2$ also differentiate to $3x^2$: constants vanish when you differentiate. So we write $+\,C$ (the constant of integration) to stand for "plus any constant":

$\int 3x^2\,dx = x^3 + C$

Read $\int \ldots\,dx$ as "integrate this with respect to $x$".

That gives the golden rule for this whole lesson: you can always check an integral by differentiating your answer. If you get back what you started with, you are right.

You already know these derivatives, from the differentiation lessons:

Now suppose the inside is $3x$ instead of $x$. By the chain rule (differentiate the outside, then multiply by the derivative of the inside):

$\frac{d}{dx}\left(e^{3x}\right) = e^{3x} \times 3 = 3e^{3x}$

So if we guess $\int e^{3x}\,dx = e^{3x}$, our guess differentiates to $3e^{3x}$, which is three times too big. Fix it by dividing by 3:

$\frac{d}{dx}\left(\tfrac{1}{3}e^{3x}\right) = \tfrac{1}{3} \times 3e^{3x} = e^{3x} \quad ✓$

The same thing happens for every function with a linear inside $ax + b$: differentiating multiplies by $a$ (the derivative of $ax + b$), so integrating has to divide by $a$. The $b$ makes no difference, because it differentiates to 0.