Integration by parts — derivation from product rule, LIATE mnemonic: Worked examples

This page is a problem gym for integration by parts. The parent note built the rule; here we hit it from every angle. Before we solve anything, we lay out a map of every kind of problem that integration by parts can throw at you — so when you meet a new integral, you already recognise which "cell" it belongs to.

Recall The one formula we lean on the whole way

$\int u d v = uv - \int v d u$ You pick $u$ (the factor you will differentiate into $d u$ ) and $d v$ (the factor you will integrate into $v$ ). The picking rule is LIATE — Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential — earliest letter becomes $u$ .

The scenario matrix

Integration by parts is not one problem — it is a family. Here is every distinct behaviour the method can produce. Every example below is tagged with the cell it fills.

Cell	Case class	Signature / what happens	Filled by
A	Poly × exponential, one pass	$d u$ makes the polynomial vanish in one step	Ex 1
B	"Hidden product" (single function)	Write $f (x) = f (x) \cdot 1$ so $d v = d x$	Ex 2
C	Repeated parts (poly degree $> 1$ )	Do parts $n$ times until poly dies	Ex 3
D	The loop (integral reappears)	Solve algebraically for $I$	Ex 4
E	Inverse-trig alone	$u = arctan x$ , leftover $\int v d u$ needs substitution	Ex 5
F	Definite integral	Carry the $[uv]_{a}^{b}$ boundary term; watch signs at the ends	Ex 6
G	Degenerate / limiting case	$\int_{0}^{1} ln x d x$ — an improper endpoint where $ln 0 \to - \infty$	Ex 7
H	Word problem (physics)	Work done by a position-dependent, oscillating force	Ex 8
I	Exam twist (wrong- $u$ trap)	Show the bad choice explodes, then rescue with LIATE	Ex 9

Signs and directions we must cover explicitly: the leading minus in $uv - \int v d u$ ; negative arguments/limits (Ex 6 runs from $0$ to $π$ where $cos$ flips sign); a limit going to $- \infty$ (Ex 7); and both signs of the loop constant (Ex 4).

Cell A — Polynomial × exponential (one pass)

Worked example Example 1 —

$\int x e^{2 x} d x$ Forecast: guess before reading — after one round of parts, will the leftover integral be simpler than the start? (Yes, because the $x$ turns into a plain constant.)

Classify with LIATE. $x$ is Algebraic (A), $e^{2 x}$ is Exponential (E). A comes before E, so $u = x$ . Why this step? We need $u$ to be the factor that simplifies when differentiated; $x \to 1$ vanishes, $e^{2 x}$ never simplifies.
Compute $d u$ and $v$ . $u = x \Rightarrow d u = d x$ ; $d v = e^{2 x} d x \Rightarrow v = \frac{1}{2} e^{2 x}$ . Why this step? $v = \int d v$ ; integrating $e^{2 x}$ needs the $\frac{1}{2}$ from the inner factor $2$ (chain-rule in reverse — see Integration by substitution).
Apply the formula. $\int x e^{2 x} d x = x \cdot \frac{1}{2} e^{2 x} - \int \frac{1}{2} e^{2 x} d x$ Why this step? Direct substitution into $uv - \int v d u$ ; keep the leading minus.
Finish the easy integral. $= \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} + C = \frac{1}{4} e^{2 x} (2 x - 1) + C$ Why this step? $\int \frac{1}{2} e^{2 x} d x = \frac{1}{4} e^{2 x}$ ; the $x$ is gone, so we're done.

Verify: differentiate $\frac{1}{4} e^{2 x} (2 x - 1)$ : product rule gives $\frac{1}{4} (2 e^{2 x} (2 x - 1) + 2 e^{2 x}) = \frac{1}{4} e^{2 x} (4 x) = x e^{2 x}$ . ✓

Cell B — The hidden product

Worked example Example 2 —

$\int arcsin x d x$ Forecast: there's only one function here — how can "by parts" apply? (Trick: multiply by $1$ .)

Manufacture a product. Write $arcsin x = arcsin x \cdot 1$ . Then $u = arcsin x$ (I in LIATE), $d v = 1 d x$ . Why this step? Inverse-trig has no elementary antiderivative on sight but a clean derivative, so it must be $u$ .
Differentiate and integrate. $d u = \frac{1}{1 - x ^{2}} d x$ , and $v = \int 1 d x = x$ . Why this step? $\frac{d}{d x} arcsin x = \frac{1}{1 - x ^{2}}$ ; the leftover $1$ integrates to $x$ .
Apply the formula. $\int arcsin x d x = x arcsin x - \int \frac{x}{1 - x ^{2}} d x$ Why this step? Substituting into $uv - \int v d u$ ; the new integral is now a plain substitution.
Substitute $w = 1 - x^{2}$ , $d w = - 2 x d x$ : $\int \frac{x}{1 - x ^{2}} d x = - \frac{1}{2} \int w^{- 1/2} d w = - 1 - x^{2}$ Why this step? The $x d x$ in the numerator is exactly (minus half of) $d w$ , so the root untangles.
Combine. $\int arcsin x d x = x arcsin x + 1 - x^{2} + C$

Verify: $\frac{d}{d x} (x arcsin x + 1 - x^{2}) = arcsin x + \frac{x}{1 - x ^{2}} - \frac{x}{1 - x ^{2}} = arcsin x$ . ✓

Cell C — Repeated parts (polynomial dies in stages)

Worked example Example 3 —

$\int x^{2} e^{- x} d x$ Forecast: the $x^{2}$ won't vanish in one pass. How many rounds until it's gone? (Two — degree drops by one each time.)

First pass. $u = x^{2}$ , $d v = e^{- x} d x \Rightarrow d u = 2 x d x$ , $v = - e^{- x}$ . Why this step? Algebraic before Exponential; integrating $e^{- x}$ gives $- e^{- x}$ (mind the minus). $\int x^{2} e^{- x} d x = - x^{2} e^{- x} + \int 2 x e^{- x} d x$
Second pass on $\int 2 x e^{- x} d x$ : $u = 2 x$ , $d v = e^{- x} d x \Rightarrow d u = 2 d x$ , $v = - e^{- x}$ . Why this step? Still a poly×exp product; repeat until the poly is a constant. $\int 2 x e^{- x} d x = - 2 x e^{- x} + \int 2 e^{- x} d x = - 2 x e^{- x} - 2 e^{- x}$
Combine both passes. $\int x^{2} e^{- x} d x = - x^{2} e^{- x} - 2 x e^{- x} - 2 e^{- x} + C = - e^{- x} (x^{2} + 2 x + 2) + C$ Why this step? Bookkeeping: add the two results, factor out $- e^{- x}$ .

This is exactly the pattern Tabular integration (DI method) mechanises — a table beats writing all this out.

Verify: $\frac{d}{d x} [- e^{- x} (x^{2} + 2 x + 2)] = e^{- x} (x^{2} + 2 x + 2) - e^{- x} (2 x + 2) = e^{- x} x^{2}$ . ✓

Cell D — The loop trick (integral comes back)

Cell E — Inverse-trig, leftover needs a substitution

Cell F — A definite integral (carry the boundary term)

For definite integrals the rule becomes $\int_{a}^{b} u d v = [uv]_{a}^{b} - \int_{a}^{b} v d u,$ where $[uv]_{a}^{b} = u (b) v (b) - u (a) v (a)$ (this is the Fundamental Theorem of Calculus applied to the $uv$ term). See Definite integrals for the boundary bookkeeping.

Worked example Example 6 —

$\int_{0}^{π} x cos x d x$ Forecast: over $[0, π]$ the cosine goes positive then negative — do you expect a positive or negative answer? (Guess, then check.)

Choose. $u = x$ (A), $d v = cos x d x \Rightarrow d u = d x$ , $v = sin x$ .
Apply with limits. $\int_{0}^{π} x cos x d x = [x sin x]_{0}^{π} - \int_{0}^{π} sin x d x$
Evaluate the boundary term. $[x sin x]_{0}^{π} = π sin π - 0 \cdot sin 0 = π \cdot 0 - 0 = 0$ . Why this step? $sin π = 0$ and the lower end is $0$ — the whole boundary term collapses. Look at the figure: the height $x sin x$ is zero at both ends.
Evaluate the remaining integral. $\int_{0}^{π} sin x d x = [- cos x]_{0}^{π} = - cos π + cos 0 = 1 + 1 = 2$ .
Combine. $\int_{0}^{π} x cos x d x = 0 - 2 = - 2$ Why negative? On $[\frac{π}{2}, π]$ the cosine is negative and $x$ is large, so the negative area outweighs the positive early area — the red region in the figure dominates.

Verify: antiderivative is $x sin x + cos x$ ; evaluate $[\cdot]_{0}^{π} = (π \cdot 0 - 1) - (0 + 1) = - 2$ . ✓

Cell G — Degenerate / limiting endpoint (improper integral)

Worked example Example 7 —

$\int_{0}^{1} ln x d x$ Forecast: $ln x \to - \infty$ as $x \to 0^{+}$ . Does the area blow up, or stay finite? (It stays finite — the spike is thin.)

Antiderivative by parts. From the parent note, $\int ln x d x = x ln x - x$ (take $u = ln x$ , $d v = d x$ ).
Treat the bad end as a limit. Because $ln x$ is undefined at $x = 0$ , define $\int_{0}^{1} ln x d x = lim_{ε \to 0^{+}} [x ln x - x]_{ε}^{1} .$ Why this step? We must never evaluate $ln 0$ directly; we sneak up on it with $ε$ (see the shrinking sliver in the figure).
Plug in the ends. At $x = 1$ : $1 \cdot 0 - 1 = - 1$ . At $x = ε$ : $ε ln ε - ε$ .
Take the limit. $ε ln ε \to 0$ as $ε \to 0^{+}$ (the $ε$ crushes the slow $- \infty$ of $ln$ ), and $ε \to 0$ . So the lower end contributes $0$ . Why this step? This is the crucial degenerate check: the $x$ factor in $x ln x$ tames the singularity.
Result. $\int_{0}^{1} ln x d x = (- 1) - 0 = - 1$

Verify: $ε \to 0^{+} lim (ε ln ε - ε) = 0$ , and the value is $- 1$ . ✓

Cell H — Word problem (physics: work by an oscillating force)

Worked example Example 8 — Work done by a spring-plus-position force

A particle moves along a line from $x = 0$ to $x = π$ metres. The force on it (in newtons) is $F (x) = x sin x$ . Work is $W = \int_{0}^{π} F (x) d x$ . Find $W$ in joules. Forecast: the force grows with $x$ but oscillates via $sin x$ ; over $[0, π]$ , $sin x \geq 0$ , so expect $W > 0$ .

Set up parts. $u = x$ (A), $d v = sin x d x \Rightarrow d u = d x$ , $v = - cos x$ . Why this step? $x$ simplifies to $1$ ; $sin x$ integrates cleanly.
Apply with limits. $W = [- x cos x]_{0}^{π} + \int_{0}^{π} cos x d x$ Why the plus? The leading term is $- \int v d u = - \int (- cos x) d x = + \int cos x d x$ .
Boundary term. $[- x cos x]_{0}^{π} = - π cos π + 0 = - π (- 1) = π$ .
Remaining integral. $\int_{0}^{π} cos x d x = [sin x]_{0}^{π} = 0 - 0 = 0$ .
Combine. $W = π + 0 = π \approx 3.14$ joules.

Verify (units + sign): force (N) × distance (m) = joules ✓; $W = π > 0$ matches the forecast since $sin x \geq 0$ on $[0, π]$ . Numerically $W = π$ . ✓

Cell I — Exam twist: the wrong- $u$ trap

Worked example Example 9 —

$\int x sec^{2} x d x$ , and why the tempting choice fails Forecast: $sec^{2} x$ is "the derivative of $tan x$ ", so integrating it is easy — should it be $u$ or $d v$ ?

The trap ( $u = sec^{2} x$ ): then $d v = x d x$ , $v = \frac{x ^{2}}{2}$ , $d u = 2 sec^{2} x tan x d x$ . The new integral $\int \frac{x ^{2}}{2} \cdot 2 sec^{2} x tan x d x$ is worse — higher power of $x$ , nastier trig. Dead end. Why it feels right: " $sec^{2}$ integrates nicely, so make it $u$ " — but $u$ is what you differentiate, and differentiating $sec^{2} x$ makes it uglier.

The fix (LIATE: A before T $\Rightarrow u = x$ ):

$u = x$ , $d v = sec^{2} x d x \Rightarrow d u = d x$ , $v = tan x$ . Why this step? $x$ simplifies to $1$ ; $sec^{2} x$ integrates to $tan x$ .
Apply. $\int x sec^{2} x d x = x tan x - \int tan x d x$
Finish. $\int tan x d x = - ln ∣ cos x ∣$ , so $\int x sec^{2} x d x = x tan x + ln ∣ cos x ∣ + C$ Why $ln ∣ cos x ∣$ ? $tan x = \frac{sin x}{cos x}$ ; the numerator is $- \frac{d}{d x} cos x$ , giving a log of the denominator.

Verify: $\frac{d}{d x} [x tan x + ln ∣ cos x ∣] = tan x + x sec^{2} x + \frac{- sin x}{cos x} = tan x + x sec^{2} x - tan x = x sec^{2} x$ . ✓

Recap of the matrix

Recall Which trick for which signature?

Poly × exp, one pass ::: differentiate the poly once (Ex 1) Single log / inverse-trig function ::: write $f (x) \cdot 1$ , so $d v = d x$ (Ex 2, 5) Poly of degree $n$ ::: repeat parts $n$ times / use Tabular integration (DI method) (Ex 3) Integral reappears ::: name it $I$ , solve algebraically (Ex 4) Definite integral ::: carry $[uv]_{a}^{b}$ , evaluate both ends (Ex 6) Singular endpoint ::: use a limit $ε \to 0^{+}$ (Ex 7) "Easy to integrate" factor ::: that's $d v$ , NOT $u$ (Ex 9)

Connections

Parent topic — the rule and its derivation.
Integration by substitution — used inside Ex 2 and Ex 5 for the leftover integral.
Definite integrals and Fundamental Theorem of Calculus — the boundary term in Ex 6–8.
Tabular integration (DI method) — shortcut for the repeated-parts case (Ex 3).
Reduction formulae — what repeated parts becomes when the power is a general $n$ .