= Integration by parts
{title2=$\int u\,dv=uv-\int v\,du$}
{wiki}
= Integrations by parts
{synonym}
<Integration by parts> transfers a <derivative> from one factor of a product to the other, while keeping an endpoint contribution. Its usefulness comes from choosing which factor becomes simpler when differentiated, and which has an accessible <antiderivative>. The <integral> is then replaced by a <boundary term> minus another <integral>.
For real <continuously differentiable functions> $u,v$ on $[a,b]$, the precise <definite integral> formula is
$$
\int_a^b u(x)v'(x)\,dx=[u(x)v(x)]_a^b-\int_a^b u'(x)v(x)\,dx,
\qquad [uv]_a^b=u(b)v(b)-u(a)v(a).
$$
In the shorter notation, $du=u'(x)\,dx$ and $dv=v'(x)\,dx$, giving $\int u\,dv=uv-\int v\,du$ for <indefinite integrals>, with a <constant of integration> understood. The <product rule> says $(uv)'=u'v+uv'$; the <fundamental theorem of calculus> turns its <integral> into the endpoint difference. Thus the formula packages a differentiation rule into a tool for <integration>.https://openstax.org/books/calculus-volume-2/pages/3-1-integration-by-parts[OpenStax, Calculus Volume 2, Section 3.1: Integration by parts]{ref}
For example, choose $u=x$ and $dv=e^x\,dx$. Then $du=dx$ and $v=e^x$, since the <exponential function> is its own <derivative>:
$$
\int_0^1 xe^x\,dx=[xe^x]_0^1-\int_0^1e^x\,dx=e-(e-1)=1.
$$
The <boundary term> is $e\approx2.718282$; the remaining <integral> is $e-1\approx1.718282$. Their difference is exactly one. The endpoint product alone would overcount, because its <derivative> includes $u'v=e^x$ as well as the desired $uv'=xe^x$.
\Image[/integration-by-parts-example.png]
{title=Integration by parts splits the accumulated product derivative into the desired integral and the remaining integral}
In the figure, the <area> under $(xe^x)'=(1+x)e^x$ is the endpoint contribution $e$. Its lower band has <area> $e-1$, and its upper band has thickness $xe^x$ and <area> one. This decomposition makes the subtraction visible. The endpoint contribution is a difference of product values; it need not equal the <area> under either factor separately.
The same choice repeatedly lowers the <degree of a polynomial> when integrating a <polynomial> times an <exponential function>, while choosing $u=\ln x$ gives $\int\ln x\,dx=x\ln x-x+C$ for $x>0$. More broadly, moving a <derivative> onto a convenient test function is the mechanism behind a <weak derivative>. Always evaluate the <boundary term>: it disappears only when the relevant endpoint values justify doing so, often through a <boundary condition>.
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