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Integral Calculus

Integration by Parts

Mathematics I 323 words Free to read

Reversing the Product Rule

Integration by parts is the integration technique that reverses the product rule. It handles integrals of products — especially a polynomial times an exponential, logarithm, or trig function — that substitution cannot touch. The formula, derived by integrating the product rule, is: udv=uvvdu.\int u\, dv = uv - \int v\, du. You split the integrand into a part uu (to differentiate) and a part dvdv (to integrate), then trade the original integral for a hopefully simpler one, vdu\int v\, du.

The art is choosing uu and dvdv wisely — you want vdu\int v\, du to be easier than what you started with. A useful guide is the mnemonic LIATE, ordering good choices for uu: Logarithmic, Inverse trig, Algebraic (polynomial), Trigonometric, Exponential. Pick uu to be whichever type appears earliest in this list, and let dvdv be the rest.

For example, xexdx\int x\, e^x\, dx: choose u=xu = x (algebraic, so du=dxdu = dx) and dv=exdxdv = e^x\, dx (so v=exv = e^x). Then xexdx=xexexdx=xexex+C.\int x e^x\, dx = x e^x - \int e^x\, dx = x e^x - e^x + C. The new integral exdx\int e^x\, dx is trivial — a good sign the choice was right. Sometimes parts must be applied twice (e.g. x2exdx\int x^2 e^x\, dx), and occasionally it produces the original integral again, which you solve algebraically.

Substitution and parts are the two workhorses: substitution for compositions, parts for products. Recognizing which structure an integral has is the first step in choosing the technique.

Common pitfall: choosing uu and dvdv badly, so the new integral vdu\int v\, du is harder than the original. The goal is to make it simpler — pick uu to be the part that gets simpler when differentiated (a polynomial eventually becomes a constant), following LIATE. Choosing u=exu = e^x in xexdx\int x e^x\, dx, for instance, leads nowhere; the right choice (u=xu = x) makes the remaining integral trivial.

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Integral Calculus