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Mathematics · Ch 11 — Integral Calculus

Bernoulli's Formula for Integration by Parts

11.7.6

Bernoulli's Formula for Integration by Parts

When a product integrand needs integration by parts applied two, three, or more times in a row — most often when one factor is xnx^n for n≥2n\ge2 — repeating the ordinary by-parts formula step by step gets long and error-prone. Bernoulli's formula packages the whole repeated process into a single alternating-sign expression.

If uu and vv are functions of xx, Bernoulli's rule states:

∫u dv=uv−u′v1+u′′v2−u′′′v3+⋯\int u\,dv=uv-u'v_1+u''v_2-u'''v_3+\cdots

where u′,u′′,u′′′,…u',u'',u''',\ldots are the successive derivatives of uu (differentiate uu again and again until it becomes 00, which happens after finitely many steps whenever uu is a polynomial), and v,v1,v2,v3,…v,v_1,v_2,v_3,\ldots are the successive integrals of dvdv (integrate dvdv once to get vv, integrate v dxv\,dx again to get v1v_1, and so on). The pattern of signs strictly alternates +,−,+,−,…+,-,+,-,\ldots, and the process terminates automatically once a derivative of uu reaches 00 (for u=xnu=x^n with nn a positive integer, this takes exactly n+1n+1 terms).

Bernoulli's formula is most advantageous exactly when u=xnu=x^n for a positive integer nn, since only then does the chain of derivatives u′,u′′,…u',u'',\ldots terminate in finitely many steps; it saves writing out nn separate applications of the ordinary by-parts formula.

Worked-style illustration (paralleling Example 11.35). For ∫x2e5x dx\int x^2e^{5x}\,dx: take u=x2u=x^2 (so u′=2xu'=2x, u′′=2u''=2, u′′′=0u'''=0) and dv=e5xdxdv=e^{5x}dx (so v=e5x5v=\dfrac{e^{5x}}5, v1=e5x25v_1=\dfrac{e^{5x}}{25}, v2=e5x125v_2=\dfrac{e^{5x}}{125}). Bernoulli's formula gives directly

∫x2e5x dx=x2⋅e5x5−2x⋅e5x25+2⋅e5x125+c=x2e5x5−2xe5x25+2e5x125+c,\int x^2e^{5x}\,dx=x^2\cdot\dfrac{e^{5x}}5-2x\cdot\dfrac{e^{5x}}{25}+2\cdot\dfrac{e^{5x}}{125}+c=\dfrac{x^2e^{5x}}5-\dfrac{2xe^{5x}}{25}+\dfrac{2e^{5x}}{125}+c,

in one line instead of two separate by-parts steps. The same table-based bookkeeping handles ∫x3cos⁡x dx\int x^3\cos x\,dx (four terms, since u=x3u=x^3 needs three differentiations to reach 00) and ∫x3e−x dx\int x^3e^{-x}\,dx equally directly. …