About integration by substitution I know how the method goes: we want to find $\int{f(g(x))g'(x)dx}$, which by the reverse chain rule equals $\int{f(u)du}$. My (maybe stupid) question comes from the integrals with the form 
$$\int{\frac{Ax+B}{(x+\alpha)^2+\beta^2}}.$$ 
The explanation given in class was that, in order to solve it, you have to substitute $x+\alpha$ with $\beta y$, but how do you know that the original fraction does actually have the structure needed to be able to use the substitution method? Don't you have to determine first if it has the form of $\int{f(g(x)g'(x)dx}$ to justify the change? 
EDIT: I overcomplicated the question, sorry. What I'm trying to say is that, most of the time, the substitution is done directly, without justifying  that the original expression has the form of $\int{f(g(x))g'(x)dx}$ (and, therefore, the reverse chain rule applies), and I was wondering why. Is it because all expressions can be put so as to acquire that form?
 A: The nice thing about the substitution $y = \frac{x + \alpha}{\beta}$ is that it's a linear substitution, and linear substitutions can always be made neatly. There is a nice shorthand using differentials that we use when making substitutions. We start by differentiating:
$$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\beta} \implies \mathrm{d}x = \beta \mathrm{d}y,$$
then we substitute into the integral:
$$\int \frac{Ax + B}{(x + \alpha)^2 + \beta^2} \mathrm{dx} = \int \frac{A(\beta y - \alpha) + B}{(\beta y)^2 + \beta^2} \beta\mathrm{dy}.$$
We can still rigidly apply the formula provided. In this case, we wish to make $g(x) = \frac{x + \alpha}{\beta}$. Then $g'(x) = \frac{1}{\beta}$. We can then write,
$$\int \frac{Ax + B}{(x + \alpha)^2 + \beta^2} \mathrm{dx} = \int \frac{Ax + B}{(x + \alpha)^2 + \beta^2} \beta g'(x) \mathrm{dx}.$$
All we have to do is find a function $f$ such that
$$f(g(x)) = f\left(\frac{x + a}{\beta}\right) = \frac{Ax + B}{(x + \alpha)^2 + \beta^2} \beta$$
for all $x$. To find such an $x$, take the above equation, and substitute into both sides $x = \beta{y} - \alpha$, giving us,
$$f(y) = \frac{A(\beta y - \alpha) + B}{(\beta y)^2 + \beta^2} \beta.$$
Using this definition of $f$, we have,
$$\int f(g(x)) g'(x) \mathrm{d}x = \int \frac{Ax + B}{(x + \alpha)^2 + \beta^2} \mathrm{dx},$$
which implies, taking $u = g(x) = \frac{x + \alpha}{\beta}$, the integral is equal to,
$$\int f(u) \mathrm{d}u = \int \frac{A(\beta u - \alpha) + B}{(\beta u)^2 + \beta^2} \beta \mathrm{d}u.$$
We get the same answer, and hopefully you can see how the steps mirror each other. It's just that the shorthand is a lot quicker to deal with.
