What happens when substitution fails? When you face $\int x e^x \, dx$ or $\int x \ln x \, dx$ and no substitution simplifies the integrand? Integration by parts is the answer. Derived from the product rule for differentiation, it transforms one integral into another — hopefully simpler — integral. It's the technique that bridges the gap between basic integration and the integrals that appear in physics, engineering, and probability theory. The choice of which part to differentiate and which to integrate is an art, and mastering it separates competent students from exceptional ones.
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Consider $\int x e^x \, dx$. If you try $u = x$ and $dv = e^x \, dx$, then $du = dx$ and $v = e^x$. Using the formula $\int u \, dv = uv - \int v \, du$, what does the integral become?
Core Content
The product rule for differentiation says:
d/dx(uv) = u'v + uv'
Rearranging: $uv' = (uv)' - u'v$
Integrating both sides:
∫ uv' dx = uv - ∫ u'v dx
In differential notation, with $u = u(x)$, $dv = v'(x)\,dx$:
∫ u dv = uv - ∫ v du
The strategy: Split the integrand into $u$ and $dv$. Differentiate $u$ to get $du$. Integrate $dv$ to get $v$. The new integral $\int v \, du$ should be simpler than the original.
When applying integration by parts, choosing which part is $u$ and which is $dv$ is crucial. The LIATE rule provides a priority order:
Choose $u$ as the function that appears earliest in LIATE. This typically makes $\int v \, du$ simpler than the original.
Example: For $\int x e^x \, dx$, LIATE says Algebraic ($x$) before Exponential ($e^x$), so $u = x$ and $dv = e^x \, dx$.
Example: For $\int x \ln x \, dx$, LIATE says Logarithmic ($\ln x$) before Algebraic ($x$), so $u = \ln x$ and $dv = x \, dx$.
Find $\int x \ln x \, dx$.
The indefinite integral.
LIATE: L before A, so u = ln x, dv = x dx
du = (1/x) dx, v = x^2/2
∫ x ln x dx = (x^2/2)ln x - ∫ (x^2/2)(1/x) dx
= (x^2/2)ln x - (1/2)∫ x dx
= (x^2/2)ln x - x^2/4 + C
$\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$.
Find $\int x \cos x \, dx$.
Answer:
$u = x$, $dv = \cos x \, dx$ → $du = dx$, $v = \sin x$.
$\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C$.
$\int u \, dv = uv - \int v \, du$
L-I-A-T-E priority for choosing $u$
$[uv]_a^b - \int_a^b v \, du$
Differentiate your answer to verify
With $u = x$ and $dv = e^x \, dx$, integration by parts gives $\int x e^x \, dx = x e^x - \int e^x \, dx = x e^x - e^x + C = e^x(x - 1) + C$. The key insight is that differentiating $x$ reduces it to $1$, making the remaining integral trivial. If we had chosen $u = e^x$ instead, we would get $\int x e^x \, dx = \frac{x^2}{2}e^x - \int \frac{x^2}{2}e^x \, dx$, which is more complicated than what we started with. The LIATE rule exists precisely to prevent this mistake — it tells us to differentiate the part that simplifies when differentiated.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Find $\int x e^{2x} \, dx$ using integration by parts. Show all working. 3 MARKS
9. Evaluate $\int_0^{\pi/2} x \sin x \, dx$. Show all working. 4 MARKS
10. The expected lifetime of a radioactive atom with decay constant $\lambda$ is $E[T] = \int_0^{\infty} t \cdot \lambda e^{-\lambda t} \, dt$. Use integration by parts to show that $E[T] = \frac{1}{\lambda}$. Explain what this means physically. 4 MARKS
1. $u = x$, $dv = e^x dx$. $\int x e^x dx = x e^x - e^x + C = e^x(x-1) + C$.
2. First: $u = x^2$, $dv = e^x dx$ → $x^2 e^x - 2\int x e^x dx$. Then apply by parts again to $\int x e^x dx$. Final: $e^x(x^2 - 2x + 2) + C$.
3. $u = \ln x$, $dv = dx$. $[x \ln x]_1^e - \int_1^e dx = e - (e - 1) = 1$.
1. Then $du = e^x dx$, $v = x^2/2$, giving $\frac{x^2}{2}e^x - \int \frac{x^2}{2}e^x dx$ — harder than original.
2. First application: $\int e^x \sin x dx = e^x \sin x - \int e^x \cos x dx$. Second: $\int e^x \cos x dx = e^x \cos x + \int e^x \sin x dx$. Substitute back and solve: $2\int e^x \sin x dx = e^x(\sin x - \cos x)$, so integral $= \frac{e^x}{2}(\sin x - \cos x) + C$.
3. Integration by parts is obtained by integrating the product rule rearrangement $uv' = (uv)' - u'v$.
Q8 (3 marks): $u = x$, $dv = e^{2x} dx$, so $du = dx$, $v = \frac{1}{2}e^{2x}$ [1]. $\int x e^{2x} dx = \frac{x}{2}e^{2x} - \frac{1}{2}\int e^{2x} dx$ [1] $= \frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C = \frac{e^{2x}}{4}(2x - 1) + C$ [1].
Q9 (4 marks): $u = x$, $dv = \sin x dx$, so $du = dx$, $v = -\cos x$ [1]. $\int_0^{\pi/2} x \sin x dx = [-x \cos x]_0^{\pi/2} + \int_0^{\pi/2} \cos x dx$ [1] $= (0 - 0) + [\sin x]_0^{\pi/2}$ [1] $= 1 - 0 = 1$ [1].
Q10 (4 marks): $u = t$, $dv = \lambda e^{-\lambda t} dt$, so $du = dt$, $v = -e^{-\lambda t}$ [1]. $E[T] = [-t e^{-\lambda t}]_0^{\infty} + \int_0^{\infty} e^{-\lambda t} dt$ [1]. The boundary term is 0 (exponential decay beats linear growth) [0.5]. $\int_0^{\infty} e^{-\lambda t} dt = [-\frac{1}{\lambda}e^{-\lambda t}]_0^{\infty} = \frac{1}{\lambda}$ [1]. So $E[T] = \frac{1}{\lambda}$ [0.5]. Physically, this means the average lifetime is inversely proportional to the decay rate — faster-decaying atoms have shorter average lives [1].
Climb platforms by applying integration by parts, choosing u and dv using LIATE, and evaluating the results. Pool: lesson 9.
Tick when you've finished all activities and checked your answers.