Intuition The one-line ideaDifferentiation has a product rule . Integration is just differentiation run backwards, so the product rule, when integrated, must give us a rule for integrals too. That rule is integration by parts . It lets us trade a hard integral ∫ u d v \int u\,dv ∫ u d v ∫ v d u \int v\,du ∫ v d u
When you integrate a product of two different "types" of functions (like x ⋅ e x x\cdot e^x x ⋅ e x x ln x x\ln x x ln x differentiate a product. So instead of inventing something new, we reverse the product rule . The genius move: integrating both sides of an identity we already trust.
Definition Integration by partsFor differentiable functions u ( x ) u(x) u ( x ) v ( x ) v(x) v ( x ) ∫ u d v = u v − ∫ v d u \int u\,dv = uv - \int v\,du ∫ u d v = uv − ∫ v d u choose one factor to be u u u d u du d u d v dv d v v v v
That's it — nothing memorized, everything built .
Intuition WHY we need a guideThe rule replaces ∫ u d v \int u\,dv ∫ u d v ∫ v d u \int v\,du ∫ v d u helps if ∫ v d u \int v\,du ∫ v d u easier . The key: pick u u u simpler when differentiated (ideally vanishing), and d v dv d v
Mnemonic LIATE — order of preference foru u u u u u first in this list:
L — Logarithmic (ln x \ln x ln x log x \log x log x I — Inverse trig (arctan x \arctan x arctan x arcsin x \arcsin x arcsin x A — Algebraic (x x x x 2 x^2 x 2 T — Trigonometric (sin x \sin x sin x cos x \cos x cos x E — Exponential (e x e^x e x a x a^x a x
Whatever is earlier = u u u later = d v dv d v Why it works: Logs and inverse-trig are awful to integrate but lovely to differentiate, so they go first as u u u d v dv d v
Worked example Example 1 —∫ x e x d x \int x\,e^x\,dx ∫ x e x d x Identify types: x x x e x e^x e x u = x u=x u = x
u = x ⇒ d u = d x u = x \Rightarrow du = dx u = x ⇒ d u = d x Why? Differentiating x x x d v = e x d x ⇒ v = e x dv = e^x dx \Rightarrow v = e^x d v = e x d x ⇒ v = e x Why? e x e^x e x
Apply ∫ u d v = u v − ∫ v d u \int u\,dv = uv - \int v\,du ∫ u d v = uv − ∫ v d u ∫ x e x d x = x e x − ∫ e x d x = x e x − e x + C = e x ( x − 1 ) + C \int x e^x dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1)+C ∫ x e x d x = x e x − ∫ e x d x = x e x − e x + C = e x ( x − 1 ) + C Why this works: the new integral ∫ e x d x \int e^x dx ∫ e x d x x x x
Check (Forecast-then-Verify): differentiate e x ( x − 1 ) e^x(x-1) e x ( x − 1 ) e x ( x − 1 ) + e x = x e x e^x(x-1)+e^x = xe^x e x ( x − 1 ) + e x = x e x
Worked example Example 2 —∫ ln x d x \int \ln x\,dx ∫ ln x d x ln x = ln x ⋅ 1 \ln x = \ln x \cdot 1 ln x = ln x ⋅ 1
u = ln x ⇒ d u = 1 x d x u = \ln x \Rightarrow du = \frac{1}{x}dx u = ln x ⇒ d u = x 1 d x Why? L is first in LIATE; also ln x \ln x ln x d v = 1 d x ⇒ v = x dv = 1\,dx \Rightarrow v = x d v = 1 d x ⇒ v = x Why? The leftover factor is 1 1 1 x x x
∫ ln x d x = x ln x − ∫ x ⋅ 1 x d x = x ln x − ∫ 1 d x = x ln x − x + C \int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C ∫ ln x d x = x ln x − ∫ x ⋅ x 1 d x = x ln x − ∫ 1 d x = x ln x − x + C Verify: d d x ( x ln x − x ) = ln x + 1 − 1 = ln x \frac{d}{dx}(x\ln x - x) = \ln x + 1 - 1 = \ln x d x d ( x ln x − x ) = ln x + 1 − 1 = ln x
Worked example Example 3 — Applying it twice:∫ x 2 cos x d x \int x^2 \cos x\,dx ∫ x 2 cos x d x u = x 2 u = x^2 u = x 2 d v = cos x d x dv = \cos x\,dx d v = cos x d x d u = 2 x d x du = 2x\,dx d u = 2 x d x v = sin x v = \sin x v = sin x ∫ x 2 cos x d x = x 2 sin x − ∫ 2 x sin x d x \int x^2\cos x\,dx = x^2\sin x - \int 2x\sin x\,dx ∫ x 2 cos x d x = x 2 sin x − ∫ 2 x sin x d x ∫ 2 x sin x d x \int 2x\sin x\,dx ∫ 2 x sin x d x do parts again (u = 2 x u=2x u = 2 x d v = sin x d x dv=\sin x\,dx d v = sin x d x ∫ 2 x sin x d x = − 2 x cos x + ∫ 2 cos x d x = − 2 x cos x + 2 sin x \int 2x\sin x\,dx = -2x\cos x + \int 2\cos x\,dx = -2x\cos x + 2\sin x ∫ 2 x sin x d x = − 2 x cos x + ∫ 2 cos x d x = − 2 x cos x + 2 sin x ∫ x 2 cos x d x = x 2 sin x + 2 x cos x − 2 sin x + C \int x^2\cos x\,dx = x^2\sin x + 2x\cos x - 2\sin x + C ∫ x 2 cos x d x = x 2 sin x + 2 x cos x − 2 sin x + C Why repeat? Each pass drops the polynomial degree by one until it vanishes.
Worked example Example 4 — The "loop" trick:∫ e x sin x d x \int e^x\sin x\,dx ∫ e x sin x d x u = sin x u=\sin x u = sin x d v = e x d x ⇒ d u = cos x d x , v = e x dv=e^x dx \Rightarrow du=\cos x\,dx,\ v=e^x d v = e x d x ⇒ d u = cos x d x , v = e x I = e x sin x − ∫ e x cos x d x I = e^x\sin x - \int e^x\cos x\,dx I = e x sin x − ∫ e x cos x d x u = cos x , d v = e x d x u=\cos x,\ dv=e^x dx u = cos x , d v = e x d x ∫ e x cos x d x = e x cos x + ∫ e x sin x d x = e x cos x + I \int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx = e^x\cos x + I ∫ e x cos x d x = e x cos x + ∫ e x sin x d x = e x cos x + I I = e x sin x − ( e x cos x + I ) ⇒ 2 I = e x ( sin x − cos x ) I = e^x\sin x - (e^x\cos x + I) \Rightarrow 2I = e^x(\sin x - \cos x) I = e x sin x − ( e x cos x + I ) ⇒ 2 I = e x ( sin x − cos x ) I = 1 2 e x ( sin x − cos x ) + C I = \tfrac{1}{2}e^x(\sin x - \cos x) + C I = 2 1 e x ( sin x − cos x ) + C Why this trick: the integral reappears, so we solve for it algebraically instead of forever.
u = e x u = e^x u = e x ∫ x e x d x \int xe^x dx ∫ x e x d x Why it feels right: e x e^x e x d v dv d v opposite . Students reverse it because "easy to integrate = make it u u u The fix: u u u differentiate . You want u u u simpler . With u = e x u=e^x u = e x d u = e x d x du=e^x dx d u = e x d x ∫ e x ⋅ x 2 2 d x \int e^x\cdot\frac{x^2}{2}dx ∫ e x ⋅ 2 x 2 d x worse . LIATE saves you: A before E ⇒ u = x u=x u = x
Common mistake Forgetting thatd v dv d v d x dx d x Why it feels right: people split "x x x e x e^x e x d x dx d x The fix: Always d v = ( factor ) d x dv = (\text{factor})\,dx d v = ( factor ) d x d x dx d x
Common mistake Dropping the+ C +C + C − ∫ v d u -\int v\,du − ∫ v d u Why it feels right: the formula has a minus and it's easy to lose it after two passes.
The fix: write the boxed formula fresh each time; track the leading − - −
Recall Feynman: explain to a 12-year-oldImagine you have to clean a really messy multiplication of two things stuck together, like a sticky candy made of caramel (x x x e x e^x e x do know how to build such a candy by gluing (that's the product rule). So you run the gluing machine backwards : you trade your messy candy for a slightly cleaner one, do that trade as many times as needed, and eventually you're left with a candy so simple it's just a plain piece you already know how to eat (integrate). LIATE is just the "which piece do I peel first" instructions so you always peel the right side.
Integration by parts formula ∫ u d v = u v − ∫ v d u \int u\,dv = uv - \int v\,du ∫ u d v = uv − ∫ v d u What identity is integration by parts derived from? The product rule
d d x ( u v ) = u v ′ + v u ′ \frac{d}{dx}(uv)=u\,v'+v\,u' d x d ( uv ) = u v ′ + v u ′ , integrated on both sides.
What does LIATE stand for? Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential — order for choosing
u u u .
In LIATE, which function becomes u u u The one appearing earlier in the list (gets differentiated).
Why pick u u u Because the rule replaces
∫ u d v \int u\,dv ∫ u d v with
∫ v d u \int v\,du ∫ v d u ; that's only helpful if
d u du d u makes the integral easier.
How do you integrate ∫ ln x d x \int \ln x\,dx ∫ ln x d x Write
ln x ⋅ 1 \ln x\cdot 1 ln x ⋅ 1 , take
u = ln x , d v = d x u=\ln x,\ dv=dx u = ln x , d v = d x →
x ln x − x + C x\ln x - x + C x ln x − x + C .
∫ x e x d x = ? \int xe^x dx = ? ∫ x e x d x = ? e x ( x − 1 ) + C e^x(x-1)+C e x ( x − 1 ) + C .
When do you use the "loop"/algebraic trick? When the original integral reappears after repeated parts (e.g.
∫ e x sin x d x \int e^x\sin x\,dx ∫ e x sin x d x ); solve algebraically.
∫ e x sin x d x = ? \int e^x\sin x\,dx = ? ∫ e x sin x d x = ? 1 2 e x ( sin x − cos x ) + C \tfrac12 e^x(\sin x-\cos x)+C 2 1 e x ( sin x − cos x ) + C .
Why must d v dv d v d x dx d x Because
v = ∫ d v v=\int dv v = ∫ d v ; the
d x dx d x marks what's being integrated.
Product rule (differentiation) — the source identity.Fundamental Theorem of Calculus — justifies ∫ d d x ( u v ) d x = u v \int \frac{d}{dx}(uv)dx = uv ∫ d x d ( uv ) d x = uv Integration by substitution — the other main technique; try it first.Reduction formulae — built by applying parts repeatedly.Tabular integration (DI method) — shortcut for polynomial × (exp/trig).Definite integrals — parts with limits: [ u v ] a b − ∫ a b v d u [uv]_a^b - \int_a^b v\,du [ uv ] a b − ∫ a b v d u
Fundamental Theorem of Calculus
Solve for wanted integral
Integration by Parts formula
Choice problem: which is u?
Pick u: simpler when differentiated
Pick dv: easy to integrate
Intuition Hinglish mein samjhoDekho, integration by parts ka asli secret yeh hai: hum koi naya magic formula nahi rat-te. Hume product rule pehle se aata hai — d d x ( u v ) = u v ′ + v u ′ \frac{d}{dx}(uv) = u\,v' + v\,u' d x d ( uv ) = u v ′ + v u ′ ∫ u d v = u v − ∫ v d u \int u\,dv = uv - \int v\,du ∫ u d v = uv − ∫ v d u swap kar rahe hain.
Ab sabse bada confusion: u u u LIATE kaam aata hai — Logarithmic, Inverse-trig, Algebraic, Trigonometric, Exponential. Jo function is list mein pehle aaye, usko u u u d v dv d v ∫ x e x d x \int x e^x dx ∫ x e x d x x x x e x e^x e x u = x u=x u = x x x x 1 1 1
Ek classic trick: ∫ ln x d x \int \ln x\,dx ∫ ln x d x ln x × 1 \ln x \times 1 ln x × 1 u = ln x u=\ln x u = ln x d v = d x dv=dx d v = d x x ln x − x + C x\ln x - x + C x ln x − x + C ∫ e x sin x d x \int e^x \sin x\,dx ∫ e x sin x d x
Yaad rakhna: u u u simple ho, d v dv d v d x dx d x − ∫ v d u -\int v\,du − ∫ v d u
