Integration by parts ($\int udv=uv-\int vdu$)

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 isIn the shorter notation, $du=u^{\prime }(x)dx$ and $dv=v^{\prime }(x)dx$, giving $\int udv=uv-\int vdu$ for indefinite integrals, with a constant of integration understood. The product rule says $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$; the fundamental theorem of calculus turns its integral into the endpoint difference. Thus the formula packages a differentiation rule into a tool for integration.^[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:The boundary term is $e\approx 2.718282$; the remaining integral is $e-1\approx 1.718282$. Their difference is exactly one. The endpoint product alone would overcount, because its derivative includes $u^{\prime }v=e^x$ as well as the desired $u{v}^{\prime }=x{e}^x$.

In the figure, the area under $(x{e}^x{)}^{\prime }=(1+x)e^x$ is the endpoint contribution $e$. Its lower band has area $e-1$, and its upper band has thickness $x{e}^x$ 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 $\int \ln xdx=x\ln x-x+C$ 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.

Define the definite integral $F(k)={\int }_0^{\pi }{x}^k\sin xdx$. For real $k>1$, two applications of integration by parts giveThe boundary term $x^{k-1}\sin x$ vanishes at both endpoints. At $k=1$, direct integration gives $F(1)=\pi$, consistent with the displayed relation; $F(-1)$ is finite because $\sin x/x\to 1$ at zero. This recurrence and $F(0)=2$ evaluate every nonnegative integer moment, with separate even and odd chains.

A boundary term is an endpoint or boundary integral left when derivatives are transferred by integration by parts. It vanishes under some boundary conditions and otherwise supplies natural boundary conditions, conserved charges, or physical surface forces.