When the integrand is a product of functions of different nature (a polynomial times an exponential, a logarithm, etc.), one uses integration by parts, which arises from the product rule for differentiation read backwards.

Property — Integration by parts

fgdx=fgfgdx.\int f'g\,dx = fg - \int fg'\,dx. Rule: ff' is integrated (f\to f), gg is differentiated (g\to g'). It must be chosen so that the “second piece” fgdx\int fg'\,dx is simpler than the original.

Example — 1elnxdx=1\int_1^e\ln x\,dx = 1

We choose f=1f=xf'=1 \to f=x; g=lnxg=1/xg=\ln x \to g'=1/x: [xlnx]1e1ex1xdx=(e10)[x]1e=e(e1)=1.\bigl[x\ln x\bigr]_1^e - \int_1^e x\cdot\frac{1}{x}\,dx = (e\cdot 1 - 0) - [x]_1^e = e-(e-1) = \boxed{1}.

Example — Wrong vs right choice: xe2xdx\int xe^{2x}\,dx

Wrong choice: f=xf'=x, g=e2xg=e^{2x} \to the second piece becomes x222e2xdx\int\dfrac{x^2}{2}\cdot 2e^{2x}\,dxmore complicated!

Right choice: f=e2xf=e2x2f'=e^{2x} \to f=\dfrac{e^{2x}}{2}; g=xg=1g=x \to g'=1: xe2x2e2x2dx=xe2x2e2x4+C.\frac{xe^{2x}}{2} - \int\frac{e^{2x}}{2}\,dx = \frac{xe^{2x}}{2} - \frac{e^{2x}}{4}+C.

The practical rule: differentiate the factor that simplifies when differentiated (the polynomial, the logarithm), integrate the one that stays “equal to itself” (the exponential).

Topics: Integrale
Concepts: Integrale per parti
Methods: Integrale parti
Skills: Calcolare · Integrare