Integrals and the Natural Log

We know, you're expecting us to begin this section by telling you what $$\int \ln x \ dx$$ equals. But that's not really what's important here. If you really must know, it equals $x\ln x - x + C$, but few professors expect you to remember that (although it can be worked out using integration by parts, a section coming up shortly, and there, it makes a good problem.)

No, in this section we would just like to reverse the formula

$$\frac{d}{dx} (\ln x) = \frac{1}{x}$$

to obtain one of the most famous formulas in all calculus:

$$\boxed{\int \frac{1}{x} \ dx = \ln {\vert x\vert} + C}$$

There are a couple of things to notice here. First, by integrating a function 1/x that appears to have nothing whatsoever to do with e , wouldn't recognize e if e bit it on the nose, we get the log base e. Just another way of saying,``e, you're mighty special."

Second, you will notice that somehow, the x in the solution picked up absolute value signs. That's because the natural log function is only defined for positive values of x, so if x were negative and we didn't have the absolute value signs, it would make no sense. Should you worry about the absolute value signs? If you're going for the A+, YES, otherwise forget about them.

Problem   Find $$\int \tan x \ dx.$$

Now $\tan x$ is just $\sin x / \cos x$.

$$\int \tan x \ dx = \int \frac{\sin x}{\cos x} \ dx .$$

This looks ripe for a u-substitution. Let's take $u = \cos x$, so $du = - \sin x dx$ and

$$\begin{eqnarray*}\int \frac{\sin x}{\cos x} \ dx & = & \\ & = & \int \frac{-1}{... ...{1}{\vert\cos x\vert} + C \\ & = & \ln {\vert\sec x \vert} + C. \end{eqnarray*}$$

Now, was that as good for you as it was for us?