Nothing new, and that is the difficulty AB & BC

Run these top to bottom and stop at the first one that gets somewhere.

1. Is it already a basic form? A power, $e^{x}$, $\sin x$, $\cos x$, $\sec^{2}x$, $\frac{1}{x}$, $\frac{1}{1+x^{2}}$. If so, write the antiderivative and stop.
2. Would algebra or an identity simplify it? Expanding, splitting a numerator, dividing, completing the square, or a trigonometric identity. Rewriting is a technique, not a preliminary.
3. Is some inside function's derivative present, up to a constant? Substitute.
4. Is it a product of two unlike things? Integrate by parts. (BC only; the AB path stops after question 3.)

The order matters. Question 2 before question 3 saves a great deal of work: $\int \frac{x^{2}+1}{x}dx$ is a one-line split, and treating it as a substitution problem wastes several minutes before failing.

Question 3 before question 4 matters for the same reason. $\int x\left(x^{2}+1\right)^{4}dx$ can be attacked by parts and it takes four applications; substitution takes one line.

And there is a fifth possibility that is a legitimate answer: some integrands have no elementary antiderivative. $\int e^{x^{2}}dx$, $\int \frac{\sin x}{x}dx$ and $\int \sqrt{1 + x^{4}}\,dx$ are not exercises with a trick you have not spotted. Recognising one saves the time that would go into hunting.

Each technique announces itself with a structure, and the structures are worth naming.

1. Substitution. A composition, and a factor that is the derivative of what is inside it. Also: any denominator whose derivative is sitting in the numerator, which gives a logarithm.
2. Parts. A product of two pieces from different families, where one of them gets simpler on differentiating. A power times an exponential, a power times a trigonometric function, or a logarithm or inverse trigonometric function on its own.
3. Divide. A rational function whose numerator's degree is at least the denominator's.
4. Complete the square. A proper rational function with an irreducible quadratic downstairs and a constant on top.
5. Identity. An even power of sine or cosine, or a $\tan^{2}$, where the half-angle or Pythagorean identity converts it to something on the basic list.

Two structures that look like the same problem and are not:

$$\int x\,e^{x^{2}}\,dx \quad\text{is a substitution}, \qquad \int x\,e^{x}\,dx \quad\text{is parts}.$$

The difference is which position the $x$ occupies. In the first it is the derivative of the inside; in the second there is no inside at all, so nothing can be substituted for.

Similarly, $\int \frac{x}{x^{2}+1}dx$ is a substitution giving a logarithm, $\int \frac{dx}{x^{2}+1}$ is an arctangent on sight, and $\int \frac{x^{2}}{x^{2}+1}dx$ needs division first. Three integrands, three methods, one symbol apart.

The step most often skipped is question 2, because the integrand does not look like it needs anything. Three families where it does.

1. Expand a product or a power. $\int \frac{(x+1)^{2}}{\sqrt{x}}dx$ becomes $\int\left(x^{3/2} + 2x^{1/2} + x^{-1/2}\right)dx$, three power-rule terms.
2. Split a numerator over a single denominator. $\int \frac{x^{2}+1}{x}dx = \int\left(x + \frac{1}{x}\right)dx$.
3. Use an identity. $\int \sin^{2}x\,dx$ is not $\frac{\sin^{3}x}{3}$; substituting $\sin^{2}x = \frac{1 - \cos 2x}{2}$ makes it two basic integrals. Likewise $\int \tan^{2}x\,dx = \int\left(\sec^{2}x - 1\right)dx = \tan x - x + C$.

The tell for an identity is an even power of a trigonometric function with no odd factor alongside. An odd power is a substitution instead: $\int \sin^{3}x\cos x\,dx$ takes $u = \sin x$ directly, because $\cos x\,dx$ is exactly $du$.

The rewriting does not have to make the integrand shorter, only recognisable. $x^{3/2} + 2x^{1/2} + x^{-1/2}$ is longer than $\frac{(x+1)^{2}}{\sqrt{x}}$ and every term of it is something you can antidifferentiate in one step.

A failed attempt is information, and what it is information about depends on how it failed.

1. A substitution leaves a stray variable. That condemns the choice of $u$, not the method. For $\int x^{2}\sqrt{x^{3}+1}\,dx$, taking $u = x^{2}$ leaves a root in $x$ and taking $u = x^{3}+1$ fits exactly, because $du = 3x^{2}dx$ is present.
2. Parts produces a harder integral. That condemns the split of $u$ and $dv$. Swap them and try again before changing method.
3. Parts produces the original integral back. That is not a failure at all. Solve for it algebraically, as in Topic 6.11.
4. Every choice fails and the integrand is one of the standard non-elementary ones. Say so and move on.

Two habits worth carrying into the exam.

Check by differentiating. It is fast, it is certain, and it works whatever technique produced the answer. Nothing else in this unit gives that.

Keep the answer in the original variable, and keep the $+ C$. Substitution and parts both introduce intermediate objects, and the final answer has to be free of them.