When am I allowed to use ln(x) when integrating functions? In Mathematics, we know the following is true:
$$\int \frac{1}{x} \space dx = \ln(x)$$
Not only that, this rule works for constants added to x:
$$\int \frac{1}{x + 1}\space dx = \ln(x + 1) + C{3}$$
$$\int \frac{1}{x + 3}\space dx = \ln(x + 3) + C$$
$$\int \frac{1}{x - 37}\space dx = \ln(x - 37) + C$$
$$\int \frac{1}{x - 42}\space dx = \ln(x - 42) + C$$
So its pretty safe to say that $$\int \frac{1}{x + a}\space dx = \ln(x + a) + C$$ But the moment I introduce $x^a$ where $a$ is not equal to 1, the model collapses. The integral of $1/x^a$ is not equal to $\ln(x^a)$. The same goes for $\cos(x)$, and $\sin(x)$, and other trig functions. 
So when are we allowed or not allowed to use the rule of $\ln(x)$ when integrating functions?
 A: It boils down to $u$-substitution (which if you haven't covered yet, you soon will).  You know that
$$ \int \frac{1}{x} dx = \ln x + C $$
(I'll not bother with absolute values here.  You should remember them on problems that you do for class, but they aren't the focus of this question.)
Now, to handle
$$ \int \frac{1}{ax+b}dx $$
we do a $u$-substitution with $u = ax+b$.  This makes $du=adx$, so the integral becomes
$$ \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \ln u + C = \frac{1}{a}\ln(ax+b) + C$$
Note the result: 
$$ \int \frac{1}{ax+b}dx = \frac{1}{a}\ln(ax+b) + C $$
Check it by taking a derivative of the right-hand side.  A chain rule was needed to take that derivative.  The $u$-substitution we did was the chain rule reversed.
A: Generally speaking, "using $\ln (x)$" as a rule or technique is unheard of.  When one speaks of techniques, they usually include integration by substitution, integration by parts, trig substitutions, partial fractions, etc. With introductory calculus in mind, $\ln |x|$ is defined as $\int \frac{1}{x} \ dx.$ This can be extended to  $\ln |u| = \int \frac{1}{u} \ du.$ Note that there are many more definitions for $\ln (x)$, but I felt this best related particularly to your examples.
For your first couple of examples, when choosing your $u$ to be the denominator, the $du$ is simply equal to $dx.$ This is what 'allows' the integrand to be evaluated to just $\ln |u|$ where $u$ is a linear expression. 
In regards to $\int \frac{1}{x^2 + a} \ dx$, this can be handled using an inverse tangent and would be evaluated to 
$$\dfrac{\operatorname{arctan}(\frac{x}{\sqrt{a}})}{\sqrt{a}} + C$$
For integrals of the form 
$$\int \frac{1}{x^n + a} \ dx$$ 
where $n \ge 3$, you will have to revert to partial fractions. For more on partial fractions, see this.
A: Perhaps I can reverse-address your question. Oftentimes (typically in optimization problems) when dealing with a positive real function $f$ it is easier to differentiate $\log f$ than $f$ itself. It's easy to check that the so-called logarithmic derivative satisfies $\frac{d}{dx} \log [f (x)] = \frac{f'(x)}{f(x)}.$ Note also that we can recover the original derivative by multiplying through with $f.$ In terms of primitives, this is the same as saying $\displaystyle\int \frac{f'(x)}{f(x)} dx = \log x + C.$ This is a general case of some of the formulae you presented and is useful in quickly evaluating many definite integrals by substitution - for instance, $\displaystyle\int \cot x $ $dx = \log |\sin x | +C$ is immediate by this formula.
