integration by parts — HSC mathematics extension 2, live

Integration by Parts

Integration by parts is the product rule read backwards. Differentiating uv gives u′v + uv′; integrate both sides and rearrange and you have ∫u dv = uv − ∫v du, which trades the integral of a product for a boundary term uv minus a new integral. The method only pays off when you choose u so that the new integral ∫v du is simpler than the one you started with: a power of x should drop a degree, or a logarithm should disappear. This is the real knowscape from knowhere, not a picture of one. Drag it. Watch what actually changes.

mathematics extension 2 · calculus · integration by partsdrag it · it is yours

why this one carries the topic.

You can't integrate what you're handed — so you rewrite it into something you can. Partial fractions splits one hostile quotient into a sum of standard logs; product-to-sum turns a product of trig functions into terms you already know how to integrate. Every technique here is the same move: reshape the integrand until it's a form you've already beaten.

The canvas writes the u, du, dv, v table for the chosen factor, then the line uv − ∫v du with the real new integrand, so each pass shows the power of x dropping (right choice of u) or climbing (wrong choice); a second mode builds the formula from the product rule.

what examiners catch — Students pick u = eˣ or u = sin x and watch the power of x climb, or drop the limits on the uv term in a definite integral. The working that earns the mark writes the u, du, dv, v table, checks that ∫v du is simpler before continuing, and evaluates the boundary term between the same limits as the integral.

five things, not forty.

Differentiate uv with the product rule, integrate both sides and rearrange: that is where ∫u dv = uv − ∫v du comes from.

Choose u as the factor that gets simpler when differentiated: a power of x loses a degree, ln x becomes 1/x.

dv must be something you can integrate at sight: choosing dv = ln x dx leaves you unable to write v.

If the new integral still has a power of x, apply parts again; ∫x²eˣ dx needs two passes before only ∫2eˣ dx is left.

When the original integral reappears, as in ∫eˣ sin x dx, call it I and solve the equation for I.

For a definite integral the uv term is evaluated between the same limits as the integral.

nothing here is a standalone fact.

knowhere maps every concept to what it rests on and what rests on it — 2 underneath this one, 0 built on top. Each one says why, in a sentence, not as an arrow on a diagram.

this conceptintegration by partsThe canvas writes the u, du, dv, v table for the chosen factor, then the line uv − ∫v du with the real new integrand, so each pass shows the power of x dropping (right choice of u) or climbing (wrong choice); a second mode builds the formula from the product rule.

sits under itintegration by substitutionbecause parts assumes substitution fluency

sits under itdifferentiating inverse circular functionsbecause Inverse trig factors are taken as u, so their derivatives are needed for du.

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why this one carries the topic.

five things, not forty.

nothing here is a standalone fact.

3 more, same treatment.

they don't read it. they run it.