Mathematics

Integration by Parts

Most textbooks hand you ∫ u dv = uv − ∫ v du and say memorise it. It actually comes straight from the product rule — something you already know.

Product rule

d/dx [u·v] = u·v′ + v·u′

You know this already. What if we integrated both sides?

Rearrange

u·v′ = d/dx [uv] − v·u′

Isolate u·v′ — one term is now a derivative, trivial to integrate.

Integrate both sides

∫ u v′ dx = uv − ∫ v u′ dx

Integrating d/dx[uv] just returns uv.

The formula

∫ u dv = uv − ∫ v du

Rename v′dx = dv and u′dx = du. That's it — IBP is the product rule, read backwards.

The geometric proof

Plot v against u. The curve divides a rectangle into two regions — ∫u dv (left) and ∫v du (below). Together they fill the rectangle uv.

∫u dv + ∫v du = uv ⟹ ∫u dv = uv − ∫v du

0.439

∫u dv

0.184

∫v du

0.255

Always: ∫u dv + ∫v du = uv — drag the point to check for yourself

LIATE — picking u and dv

Choose u from the highest category present. The function that is hardest to integrate becomes u — differentiate it away.

Logarithms

ln x, log₂x

u ↑

Inverse trig

arcsin x, arctan x

u ↑

Algebraic

x², xⁿ, polynomials

u ↑

Trig

sin x, cos x

dv ↓

Exponential

eˣ, e^(kx)

dv ↓

Worked example — ∫ x eˣ dx: x is Algebraic, eˣ is Exponential — A before E, so u = x.

u = x → du = dx dv = eˣ dx → v = eˣ ∫ x eˣ dx = x eˣ − ∫ eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C

∫ ln x dx: write it as ∫ ln x · 1 dx. L comes first, so u = ln x, dv = 1·dx.

u = ln x → du = (1/x) dx dv = dx → v = x = x ln x − ∫ x · (1/x) dx = x ln x − x + C

The cyclic trick

∫ eˣ sin x dx can't be finished by IBP in the usual way. Apply it twice — the original integral returns, and you solve for it like an equation.

Let I = ∫ eˣ sin x dx IBP ×1 (u = sin x, dv = eˣ dx): I = eˣ sin x − ∫ eˣ cos x dx IBP ×2 (u = cos x, dv = eˣ dx): ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx So: I = eˣ sin x − eˣ cos x − I 2I = eˣ(sin x − cos x)

Don't panic when I reappears on both sides — that's the trick, not a mistake. Solve for I algebraically:

∫ eˣ sin x dx = eˣ(sin x − cos x) / 2 + C

Always add +C. And never switch your u/dv choice between the two applications.

Quiz

6 questions · Covers everything above.

Question 1

Integration by parts is derived from which differentiation rule?

Question 2

For ∫ x ln x dx, which should be u according to LIATE?

Question 3

You apply IBP to ∫eˣcos x dx and the original integral reappears. You should:

Question 4

What is ∫ ln x dx?

Question 5

To evaluate ∫ x² eˣ dx, you need to apply IBP:

Question 6

Which integral CANNOT be solved by IBP alone?