The chain rule, run backwards AB & BC

Look for a composition whose inside has its derivative present as a factor, up to a constant. In $\int 2x\left(x^{2} + 1\right)^{5}dx$ the inside is $x^{2} + 1$, and $2x$ is exactly its derivative.

1. Set $u$ to the inside. $u = x^{2} + 1$.
2. Differentiate. $\frac{du}{dx} = 2x$, so $du = 2x\,dx$.
3. Rewrite. The integral becomes $\int u^{5}\,du$, with no $x$ anywhere.
4. Antidifferentiate. $\frac{u^{6}}{6} + C$.
5. Put $x$ back. $\frac{\left(x^{2} + 1\right)^{6}}{6} + C$.

Step 3 is the test the method has to pass. If any $x$ survives, the substitution has not worked, and continuing anyway produces an expression that mixes two variables and means nothing. The $dx$ is replaced by $du$, not simply deleted.

Step 5 is the one that gets skipped on indefinite integrals. An answer in terms of $u$ is not an answer to a question posed in terms of $x$.

The derivative of the inside rarely appears with exactly the right coefficient, and that is fine. In $\int x\left(x^{2} + 1\right)^{4}dx$ we still take $u = x^{2}+1$ and $du = 2x\,dx$, so $x\,dx = \frac{1}{2}\,du$ and

$$\int x\left(x^{2}+1\right)^{4}dx = \frac{1}{2}\int u^{4}\,du = \frac{u^{5}}{10} + C = \frac{\left(x^{2}+1\right)^{5}}{10} + C.$$

The $\frac{1}{2}$ is legal because constants pass through an integral freely, by the constant-multiple rule from Topic 6.6.

What is not legal is supplying a missing variable factor the same way. $\int \left(x^{2}+1\right)^{4}dx$ has no $x$ available, and writing $\frac{1}{2x}\int u^{4}du$ moves an expression containing $x$ across the integral sign, which the constant-multiple rule does not permit. That integral is a different problem: expand it, or use another technique.

So the working test for whether a substitution fits is a single question: after pulling out constants, is what remains exactly $du$?

1. $\int x^{2}\sqrt{x^{3}+1}\;dx$ with $u = x^{3}+1$: $du = 3x^{2}dx$, so $x^{2}dx = \frac{1}{3}du$. It fits.
2. $\int x\sqrt{x^{3}+1}\;dx$ with the same $u$: we have $x\,dx$ and we need $x^{2}dx$. It does not fit, and no constant repairs it.
3. $\int \cos(3x)\,dx$ with $u = 3x$: $du = 3\,dx$, so $dx = \frac{1}{3}du$ and the result is $\frac{1}{3}\sin(3x) + C$. This is the linear-inside rule from Topic 6.8, and substitution is where that rule comes from.

When limits are attached they belong to $x$, and the moment the variable changes they stop being meaningful. There are exactly two correct ways to finish, and mixing them is the error.

1. Convert the limits. Push each $x$-limit through $u$, then evaluate entirely in $u$ and stop. For $\int_{0}^{2} 2x\left(x^{2}+1\right)^{4}dx$ with $u = x^{2}+1$: $x = 0$ gives $u = 1$, $x = 2$ gives $u = 5$, so it becomes $\int_{1}^{5} u^{4}\,du = \frac{5^{5} - 1}{5} = 624.8$. No substituting back, and no need for it.
2. Convert back first. Treat it as an indefinite integral, return to $x$, and then use the original limits $0$ and $2$.

Both give the same number. What produces a wrong one is doing neither, which means evaluating a $u$-antiderivative at $x$-limits, or doing both, which means converting the limits and then also substituting back.

The first of those is the dangerous one, because nothing about the result looks wrong. $\left[\frac{u^{5}}{5}\right]$ evaluated at $0$ and $2$ gives $6.4$, a perfectly ordinary number a hundred times too small, with no error message anywhere. The habit that prevents it: the moment you write $u$, rewrite the limits as $u$-limits in the same line.

Converted limits can come out in either order, and they are not required to increase. If $u$ is decreasing on the interval the upper limit will be the smaller number, and that is correct rather than something to fix by swapping, which would negate the answer.

A few patterns cover almost every substitution on the exam.

1. Inside a power, a root, an exponential or a trigonometric function. $u$ is what sits inside.
2. A denominator. $\int \frac{2x}{x^{2}+1}dx$ takes $u = x^{2}+1$ and becomes $\int \frac{du}{u} = \ln|u| + C$.
3. A logarithm with its own reciprocal alongside. $\int \frac{\ln x}{x}dx$ takes $u = \ln x$, since $du = \frac{1}{x}dx$ is exactly the rest of the integrand.
4. A trigonometric product. $\int \sin x\cos x\,dx$ takes $u = \sin x$, with $du = \cos x\,dx$. Taking $u = \cos x$ also works and carries a minus sign, and both answers are correct despite looking different.

Two closing habits.

If the first choice of $u$ fails, try the other candidate before abandoning the method. A failed substitution is evidence about that choice, not about the problem.

If no choice fits, it is not a substitution problem. $\int \left(x^{2}+1\right)^{4}dx$ wants expansion; $\int x e^{x}dx$ wants integration by parts, in Topic 6.11. Forcing a substitution that does not fit is how a stray $x$ ends up inside a $u$-integral, and everything after that point is unrecoverable.