Trade one integral for an easier one BC only

Start from the product rule and integrate both sides:

$$\frac{d}{dx}\left[u\,v\right] = u'v + uv' \quad\implies\quad uv = \int u'v\,dx + \int uv'\,dx.$$

Move one integral across, and in differential notation that is

$$\int u\,dv = uv - \int v\,du.$$

Three things are worth noticing about the statement itself before any example.

1. The minus sign is structural. It is what moving a term across an equals sign produced, so $uv + \int v\,du$ is not a variant, it is wrong.
2. Nothing has been evaluated. There is still an integral on the right. The method has succeeded only if that one is easier than the one you started with.
3. Four objects have to be written down: $u$, $dv$, $du$, and $v$. Skipping the table is where lost factors come from.

The standard example: $\int x e^{x}\,dx$ with $u = x$ and $dv = e^{x}dx$, so $du = dx$ and $v = e^{x}$:

$$\int x e^{x}\,dx = x e^{x} - \int e^{x}\,dx = x e^{x} - e^{x} + C.$$

The $x$ disappeared because differentiating it produced $1$. That is the whole mechanism.

Two conditions, and both have to hold.

1. $u$ should get simpler when differentiated. A power drops a degree; a logarithm or an inverse trigonometric function turns into something algebraic.
2. $dv$ should be something you can integrate. If you cannot produce $v$, the choice is unusable no matter how promising $u$ looked.

The mnemonic LIATE orders the usual candidates for $u$: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. Whichever comes first in that list is usually the right $u$. It is a heuristic, not a theorem, and the two conditions above are what it is approximating.

Watch the wrong choice fail on the same integral. Taking $u = e^{x}$ and $dv = x\,dx$ gives $du = e^{x}dx$ and $v = \frac{x^{2}}{2}$, so

$$\int x e^{x}\,dx = \frac{x^{2}e^{x}}{2} - \int \frac{x^{2}e^{x}}{2}\,dx.$$

Everything there is correct, and the new integral has $x^{2}$ where the old one had $x$. The power went the wrong way. A worse integral is a verdict on the choice, not on the method, and the response is to swap $u$ and $dv$ rather than to abandon parts.

One case looks like it has only one factor and does not: $\int \ln x\,dx$. Take $u = \ln x$ and $dv = dx$, so $v = x$ and

$$\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - x + C.$$

The same trick handles $\int \arctan x\,dx$ and $\int \arcsin x\,dx$, which have no other route at this level.

If $u$ is $x^{2}$, one application drops it to $x$ and leaves an integral that still needs parts. Apply it again.

$$\int x^{2}e^{x}\,dx = x^{2}e^{x} - 2\int x e^{x}\,dx = x^{2}e^{x} - 2\left(x e^{x} - e^{x}\right) + C.$$

Two failures live in that second line.

1. Stopping after one pass and reporting $x^{2}e^{x} - 2\int x e^{x}dx$ as the answer. An integral sign in a final answer means the work is unfinished.
2. Losing the bracket. The $-2$ multiplies both terms, so the expansion is $x^{2}e^{x} - 2xe^{x} + 2e^{x} + C$. Distributing over only the first term is the single most common slip across two passes.

There is also a case where parts never terminates and still finishes. For $I = \int e^{x}\sin x\,dx$, two applications return $I = e^{x}\sin x - e^{x}\cos x - I$, and solving algebraically gives $I = \frac{e^{x}\left(\sin x - \cos x\right)}{2} + C$. When the original integral reappears, treat it as an unknown rather than as a dead end. It has to reappear with a coefficient other than $+1$, which happens when the two applications keep the same choice of family for $u$.

With limits attached, every piece carries them:

$$\int_{a}^{b} u\,dv = \Big[uv\Big]_{a}^{b} - \int_{a}^{b} v\,du.$$

The boundary term $uv$ is evaluated at both limits and becomes a number, and the remaining integral keeps the same limits. For $\int_{0}^{1} x e^{x}\,dx$:

$$\Big[x e^{x}\Big]_{0}^{1} - \int_{0}^{1} e^{x}\,dx = \left(e - 0\right) - \left(e - 1\right) = 1.$$

Three places this goes wrong, all of them bookkeeping rather than calculus.

1. Leaving $uv$ as a function and evaluating only at the end. It works if the algebra is kept straight and it is much easier to evaluate it in place.
2. Forgetting that the lower limit contributes to $uv$ too. Here it happens to contribute $0$, which is exactly the sort of coincidence that builds a bad habit.
3. Writing $+ C$. Definite integrals do not carry one, and here there are two sources of temptation, since parts produces an intermediate indefinite antiderivative $v$.

On $v$: any antiderivative of $dv$ will do, and the constant cancels, so take the simplest one. For $dv = dx$ take $v = x$ rather than $x + 5$.