Figure 5.3: Geometric Representation of Integration by Parts: $uv = \int u\,dv + \int v\,du$
1. Mathematical Foundations of Integration by Parts
Integration by parts is the continuous antiderivative analogue of the product rule for differentiation. Given two continuously differentiable functions $u(x)$ and $v(x)$, recall that:
$$\frac{d}{dx}[u(x)v(x)] = u(x)\frac{dv}{dx} + v(x)\frac{du}{dx}$$Rearranging and integrating both sides with respect to $x$ yields the standard indefinite formula:
$$\int u(x)\frac{dv}{dx}\,dx = u(x)v(x) - \int v(x)\frac{du}{dx}\,dx \quad \iff \quad \int u\,dv = uv - \int v\,du$$For definite integrals spanning $[a, b]$, the boundary term must be evaluated at both endpoints before subtracting the remaining integral:
$$\int_a^b u\,dv = \left[ u(x)v(x) \right]_a^b - \int_a^b v\,du = \left( u(b)v(b) - u(a)v(a) \right) - \int_a^b v\,du$$Examiner Warning: Boundary Evaluation Timing
Never wait until after evaluating $\int_a^b v\,du$ to substitute limits into $uv$. Calculate the numerical value $[uv]_a^b$ immediately to prevent minus-sign distribution errors!
2. Strategic Selection of $u$ via the LIATE Hierarchy
The goal of integration by parts is to transform $\int u\,dv$ into a new integral $\int v\,du$ that is strictly simpler to integrate. The choice of $u$ is prioritized by the classic acronym LIATE:
| Priority | Function Category | Representative Examples | Mathematical Rationale |
|---|---|---|---|
| 1. L | Logarithmic | $\ln x, \log_a(x), \ln(x^2 + 1)$ | Differentiating eliminates transcendentals: $\frac{d}{dx}\ln x = \frac{1}{x}$ (rational). |
| 2. I | Inverse Trigonometric | $\arcsin x, \arccos x, \arctan x$ | Derivatives are algebraic: $\frac{1}{\sqrt{1-x^2}}$ or $\frac{1}{1+x^2}$, enabling cancellation. |
| 3. A | Algebraic / Polynomial | $x^n, ax + b, 3x^2 - 5$ | Differentiation reduces polynomial degree ($x^n \to n x^{n-1}$), eventually reaching a constant. |
| 4. T | Trigonometric | $\sin(kx), \cos(kx), \sec^2(kx)$ | Tolerates continuous integration without increasing complexity ($\int \sin(kx)dx = -\frac{1}{k}\cos(kx)$). |
| 5. E | Exponential | $e^{kx}, a^x, 2^{3x}$ | Easiest to integrate; antiderivative is identical up to a constant scalar factor $\frac{1}{k}$. |
3. Specialized Integration by Parts Techniques
A. The "Hidden Factor 1" Method for Solitary Functions
Integrals of single logarithmic or inverse trigonometric functions (e.g., $\int \ln x\,dx$, $\int \arctan x\,dx$, $\int (\ln x)^2\,dx$) contain only one obvious factor. To apply integration by parts, insert a dummy algebraic factor of $1$:
$$\int f(x)\,dx = \int f(x) \cdot 1\,dx \implies u = f(x), \quad dv = 1\,dx \implies du = f'(x)dx, \quad v = x$$ $$\int f(x)\,dx = x f(x) - \int x f'(x)\,dx$$Because $f'(x)$ is algebraic, the product $x f'(x)$ simplifies immediately into an elementary rational or radical expression.
B. Repeated Parts and the Tabular (DI) Method
When integrating a polynomial factor $P(x)$ multiplied by an exponential or sine/cosine factor (e.g., $\int x^3 e^{2x}dx$ or $\int x^2 \cos(3x)dx$), repeated integration by parts is required. The DI Method streamlines the algebra:
- Column D (Differentiate): Differentiate $P(x)$ repeatedly until reaching $0$.
- Column I (Integrate): Integrate the transcendental term $g(x)$ an equal number of times.
- Sign Column: Alternate signs starting with $+$ on the first row: $+ , -, +, -, \dots$
- Assembly: Multiply each entry in row $k$ of Column D by row $k+1$ of Column I, applying the alternating sign.
IB Exam Scoring Advice: Tabular Working
In IB DP AA HL Paper 1 and Paper 2 markschemes, examiners award method marks [M1] specifically for writing down the parts formula or setting up $u, dv, du, v$. If you use the tabular layout as scratchwork, always write out the formal first step $uv - \int v\,du$ to ensure you capture every mark!
C. Cyclic (Looping) Integrals
When both functions repeat under differentiation/integration (such as $\int e^{ax}\cos(bx)\,dx$ or $\int e^{ax}\sin(bx)\,dx$), applying parts twice produces a multiple of the original integral. Define $I$ as the initial integral:
$$I = \int e^{ax} \cos(bx)\,dx$$After two successive applications of parts (maintaining the same priority convention for $u$), we obtain:
$$I = (\text{boundary terms}) - \frac{b^2}{a^2} I$$Bring the term containing $I$ to the left-hand side and solve algebraically:
$$\left(1 + \frac{b^2}{a^2}\right) I = (\text{boundary terms}) \implies I = \frac{a^2}{a^2 + b^2}(\text{boundary terms}) + C$$D. Reduction Formulas
In higher-tariff Section B questions, integrals containing an integer parameter $n \ge 1$ (such as $I_n = \int x^n e^{-x} dx$ or $I_n = \int (\ln x)^n dx$) are evaluated using a recurrence relation:
$$I_n = \int u_n \,dv = \left[ u_n v \right] - \int v\,du_n = g(x, n) + c_n I_{n-1}$$Evaluating the base case $I_0$ or $I_1$ allows iterative evaluation for any arbitrary integer power $n$.
4. Integration via Partial Fraction Decompositions
A rational function $\frac{P(x)}{Q(x)}$ can only be decomposed directly if it is strictly proper ($\text{deg}(P) < \text{deg}(Q)$).
Step 1: The Proper Fraction Prerequisite (Polynomial Long Division)
If $\text{deg}(P) \ge \text{deg}(Q)$, you must perform polynomial long division first to express the integrand as a polynomial quotient plus a proper remainder:
$$\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)} \quad \text{where } \text{deg}(R) < \text{deg}(Q)$$Step 2: Canonical Denominator Decompositions for IB AA HL
| Denominator Factor Type | Decomposition Form | Standard Integral Antiderivative |
|---|---|---|
| Distinct Linear Factors | $\frac{A}{ax + b} + \frac{B}{cx + d}$ | $\frac{A}{a}\ln|ax + b| + \frac{B}{c}\ln|cx + d|$ |
| Repeated Linear Factor | $\frac{A}{ax + b} + \frac{B}{(ax + b)^2}$ | $\frac{A}{a}\ln|ax + b| - \frac{B}{a(ax + b)}$ |
| Irreducible Quadratic $(x^2 + a^2)$ | $\frac{Bx + D}{x^2 + a^2}$ | Split: $\frac{B}{2}\ln(x^2 + a^2) + \frac{D}{a}\arctan\left(\frac{x}{a}\right)$ |
Worked Examples
Worked Example 1: Solitary Inverse Function Definite Integral (Paper 1 Standard)
Evaluate the exact value of $\int_0^1 \arctan(x)\,dx$, expressing your answer in the form $\frac{\pi}{a} - \ln\sqrt{b}$ where $a, b \in \mathbb{Z}^+$.
Solution:
1. Insert dummy factor $1$ and set up integration by parts via LIATE (I precedes A):
$$u = \arctan(x) \implies du = \frac{1}{1 + x^2}\,dx \quad \text{[M1]}$$ $$dv = 1\,dx \implies v = x \quad \text{[A1]}$$2. Apply the definite integration by parts formula $\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du$:
$$\int_0^1 \arctan(x)\,dx = \left[ x\arctan(x) \right]_0^1 - \int_0^1 \frac{x}{1 + x^2}\,dx \quad \text{[M1]}$$3. Evaluate boundary term:
$$\left[ x\arctan(x) \right]_0^1 = (1 \cdot \arctan(1)) - (0 \cdot \arctan(0)) = 1 \cdot \frac{\pi}{4} - 0 = \frac{\pi}{4} \quad \text{[A1]}$$4. Evaluate the remaining rational integral using recognition / substitution ($\int \frac{f'(x)}{f(x)}dx = \ln|f(x)|$):
$$\int_0^1 \frac{x}{1 + x^2}\,dx = \frac{1}{2} \int_0^1 \frac{2x}{1 + x^2}\,dx = \frac{1}{2}\left[ \ln(1 + x^2) \right]_0^1 \quad \text{[M1]}$$ $$= \frac{1}{2}(\ln(2) - \ln(1)) = \frac{1}{2}\ln 2 = \ln\left(2^{1/2}\right) = \ln\sqrt{2} \quad \text{[A1]}$$5. Combine results:
$$\int_0^1 \arctan(x)\,dx = \frac{\pi}{4} - \ln\sqrt{2} \quad \text{[A1]}$$Examiner Note: Here $a = 4$ and $b = 2$. Writing $\frac{1}{2}\ln 2$ without converting to $\ln\sqrt{2}$ would forfeit the final accuracy mark when a specific form is demanded!
Worked Example 2: Cyclic Looping Integral (Paper 1 Standard)
Find $\int e^{3x} \sin(2x)\,dx$.
Solution:
Let $I = \int e^{3x} \sin(2x)\,dx$. Using LIATE, choose Trigonometric for $u$ and Exponential for $dv$:
$$u = \sin(2x) \implies du = 2\cos(2x)\,dx \quad \text{[M1]}$$ $$dv = e^{3x}\,dx \implies v = \frac{1}{3}e^{3x} \quad \text{[A1]}$$First application of integration by parts:
$$I = \frac{1}{3}e^{3x}\sin(2x) - \int \frac{1}{3}e^{3x} \cdot 2\cos(2x)\,dx = \frac{1}{3}e^{3x}\sin(2x) - \frac{2}{3}\int e^{3x}\cos(2x)\,dx \quad \text{[A1]}$$Apply parts a second time to the new integral $J = \int e^{3x}\cos(2x)\,dx$, maintaining identical priority:
$$u_1 = \cos(2x) \implies du_1 = -2\sin(2x)\,dx$$ $$dv_1 = e^{3x}\,dx \implies v_1 = \frac{1}{3}e^{3x} \quad \text{[M1]}$$ $$J = \frac{1}{3}e^{3x}\cos(2x) - \int \frac{1}{3}e^{3x}(-2\sin(2x))\,dx = \frac{1}{3}e^{3x}\cos(2x) + \frac{2}{3}\int e^{3x}\sin(2x)\,dx = \frac{1}{3}e^{3x}\cos(2x) + \frac{2}{3}I \quad \text{[A1]}$$Substitute $J$ back into the expression for $I$:
$$I = \frac{1}{3}e^{3x}\sin(2x) - \frac{2}{3}\left( \frac{1}{3}e^{3x}\cos(2x) + \frac{2}{3}I \right) \quad \text{[M1]}$$ $$I = \frac{1}{3}e^{3x}\sin(2x) - \frac{2}{9}e^{3x}\cos(2x) - \frac{4}{9}I$$Collect all terms in $I$ on the left-hand side:
$$\left(1 + \frac{4}{9}\right)I = \frac{13}{9}I = \frac{e^{3x}}{9}\left( 3\sin(2x) - 2\cos(2x) \right) \quad \text{[M1]}$$Multiply both sides by $\frac{9}{13}$ and add the constant of integration $C$:
$$I = \frac{e^{3x}}{13}\left( 3\sin(2x) - 2\cos(2x) \right) + C \quad \text{[A1]}$$Worked Example 3: Improper Rational Function & Partial Fractions (Paper 2 Standard)
Consider $f(x) = \frac{2x^3 - 3x^2 - 3x + 8}{x^2 - 1}$.
(a) Use polynomial long division to express $f(x)$ in the form $Ax + B + \frac{Cx + D}{x^2 - 1}$. [3 marks]
(b) Decompose $\frac{Cx + D}{x^2 - 1}$ into partial fractions. [3 marks]
(c) Hence evaluate the exact value of $\int_2^3 f(x)\,dx$. [4 marks]
Solution:
(a) Polynomial Long Division:
Divide $2x^3 - 3x^2 - 3x + 8$ by $x^2 - 1$:
$$\begin{array}{r} 2x - 3\phantom{)} \\ x^2 - 1{\overline{\smash{\big)}\,2x^3 - 3x^2 - 3x + 8\phantom{)}}} \\ \underline{-(2x^3\phantom{- 3x^2} - 2x)\phantom{+8)}} \\ -3x^2 - x + 8\phantom{)} \\ \underline{-(-3x^2\phantom{- x} + 3)\phantom{)}} \\ -x + 5\phantom{)} \end{array}$$ $$\implies f(x) = 2x - 3 + \frac{-x + 5}{x^2 - 1} \quad \text{[M1][A1][A1]}$$(b) Partial Fractions:
$$\frac{5 - x}{(x - 1)(x + 1)} = \frac{P}{x - 1} + \frac{Q}{x + 1} \quad \text{[M1]}$$ $$5 - x = P(x + 1) + Q(x - 1)$$Substitute $x = 1$: $5 - 1 = P(2) \implies 4 = 2P \implies P = 2$. [A1]
Substitute $x = -1$: $5 - (-1) = Q(-2) \implies 6 = -2Q \implies Q = -3$. [A1]
(c) Definite Integration:
$$\int_2^3 f(x)\,dx = \int_2^3 \left( 2x - 3 + \frac{2}{x - 1} - \frac{3}{x + 1} \right)\,dx \quad \text{[M1]}$$ $$= \left[ x^2 - 3x + 2\ln|x - 1| - 3\ln|x + 1| \right]_2^3 \quad \text{[A1]}$$Substitute upper limit $x = 3$:
$$(3^2 - 3(3) + 2\ln 2 - 3\ln 4) = 0 + 2\ln 2 - 3\ln(2^2) = 2\ln 2 - 6\ln 2 = -4\ln 2$$Substitute lower limit $x = 2$:
$$(2^2 - 3(2) + 2\ln 1 - 3\ln 3) = 4 - 6 + 0 - 3\ln 3 = -2 - 3\ln 3$$Subtract lower from upper: [M1]
$$(-4\ln 2) - (-2 - 3\ln 3) = 2 + 3\ln 3 - 4\ln 2 = 2 + \ln\left(\frac{3^3}{2^4}\right) = 2 + \ln\left(\frac{27}{16}\right) \quad \text{[A1]}$$Worked Example 4: Inductive Reduction Formula (Paper 3 / Section B Standard)
Let $I_n = \int_1^e x(\ln x)^n\,dx$ for $n \in \mathbb{N}$.
(a) Use integration by parts to prove that $I_n = \frac{e^2}{2} - \frac{n}{2}I_{n-1}$ for $n \ge 1$. [4 marks]
(b) Evaluate $I_0$. [1 mark]
(c) Hence determine the exact value of $I_2$. [3 marks]
Solution:
(a) Set up integration by parts with $u = (\ln x)^n$ (Logarithmic) and $dv = x\,dx$ (Algebraic):
$$u = (\ln x)^n \implies du = n(\ln x)^{n-1} \cdot \frac{1}{x}\,dx \quad \text{[M1]}$$ $$dv = x\,dx \implies v = \frac{x^2}{2} \quad \text{[A1]}$$Apply the parts formula:
$$I_n = \left[ \frac{x^2}{2}(\ln x)^n \right]_1^e - \int_1^e \frac{x^2}{2} \cdot \frac{n(\ln x)^{n-1}}{x}\,dx \quad \text{[M1]}$$Evaluate boundary term: at $x = e$, $\frac{e^2}{2}(\ln e)^n = \frac{e^2}{2}(1) = \frac{e^2}{2}$. At $x = 1$, $\ln 1 = 0$.
Simplify the integrand: $\frac{x^2}{2x} = \frac{x}{2}$.
Recognize the definition of $I_{n-1}$:
$$I_n = \frac{e^2}{2} - \frac{n}{2}I_{n-1} \quad \text{[AG]}$$(b) Base case $I_0$:
$$I_0 = \int_1^e x(\ln x)^0\,dx = \int_1^e x\,dx = \left[ \frac{x^2}{2} \right]_1^e = \frac{e^2 - 1}{2} \quad \text{[A1]}$$(c) Evaluation of $I_2$:
Using the recurrence for $n = 1$: [M1]
$$I_1 = \frac{e^2}{2} - \frac{1}{2}I_0 = \frac{e^2}{2} - \frac{1}{2}\left(\frac{e^2 - 1}{2}\right) = \frac{e^2}{2} - \frac{e^2 - 1}{4} = \frac{e^2 + 1}{4} \quad \text{[A1]}$$Using the recurrence for $n = 2$:
$$I_2 = \frac{e^2}{2} - \frac{2}{2}I_1 = \frac{e^2}{2} - I_1 = \frac{e^2}{2} - \frac{e^2 + 1}{4} = \frac{2e^2 - (e^2 + 1)}{4} = \frac{e^2 - 1}{4} \quad \text{[A1]}$$Worked Example 5: Trigonometric Substitution for Irrational Integrands (Paper 1 Standard)
Find the exact value of $\int_0^1 \frac{x^2}{\sqrt{4 - x^2}}\,dx$. [6 marks]
Solution:
Step 1: Trigonometric Substitution
Form $\sqrt{a^2 - x^2}$ with $a = 2$. Let $x = 2\sin\theta \implies dx = 2\cos\theta\,d\theta$ [M1].
Transform the radical: $\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4\cos^2\theta} = 2\cos\theta$ [A1].
Transform limits:
When $x = 0 \implies 2\sin\theta = 0 \implies \theta = 0$.
When $x = 1 \implies 2\sin\theta = 1 \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}$ [A1].
Step 2: Substitute and Simplify
$$\int_0^{\pi/6} \frac{(2\sin\theta)^2}{2\cos\theta} \cdot 2\cos\theta\,d\theta = \int_0^{\pi/6} 4\sin^2\theta\,d\theta \quad [M1]$$Step 3: Double-Angle Power Reduction
Use identity $\sin^2\theta = \frac{1 - \cos(2\theta)}{2}$:
$$= 4 \int_0^{\pi/6} \frac{1 - \cos(2\theta)}{2}\,d\theta = 2 \int_0^{\pi/6} (1 - \cos(2\theta))\,d\theta \quad [M1]$$ $$= 2\left[ \theta - \frac{1}{2}\sin(2\theta) \right]_0^{\pi/6} = 2\left( \frac{\pi}{6} - \frac{1}{2}\sin\left(\frac{\pi}{3}\right) \right) - 0$$ $$= 2\left( \frac{\pi}{6} - \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) \right) = \frac{\pi}{3} - \frac{\sqrt{3}}{2} \quad [A1]$$Examiner Commentary: Remember to change limits directly to $\theta$ when using trigonometric substitution. This avoids the time-consuming and error-prone back-substitution into $x$!
Practice Exercises (10 Tiered Questions)
Complete these 10 exam-style questions spanning core techniques to advanced HL challenge problems. Open each question to inspect the full step-by-step IB markscheme.
Find $\int x \cos(3x)\,dx$.
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LIATE Choice: Algebraic $u = x$, Trigonometric $dv = \cos(3x)\,dx$.
$$du = dx, \quad v = \frac{1}{3}\sin(3x) \quad \text{[M1]}$$ $$\int x \cos(3x)\,dx = \frac{1}{3}x\sin(3x) - \int \frac{1}{3}\sin(3x)\,dx \quad \text{[M1]}$$ $$= \frac{1}{3}x\sin(3x) - \frac{1}{3}\left( -\frac{1}{3}\cos(3x) \right) + C$$ $$= \frac{1}{3}x\sin(3x) + \frac{1}{9}\cos(3x) + C \quad \text{[A1]}$$Examiner Tip: Always check your signs when integrating $\sin(3x) \to -\frac{1}{3}\cos(3x)$. Two consecutive negatives yield a positive.
Evaluate $\int x^3 \ln x\,dx$.
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LIATE Choice: Logarithmic takes priority over Algebraic: $u = \ln x, dv = x^3\,dx$.
$$du = \frac{1}{x}dx, \quad v = \frac{x^4}{4} \quad \text{[M1]}$$ $$\int x^3 \ln x\,dx = \frac{x^4}{4}\ln x - \int \frac{x^4}{4} \cdot \frac{1}{x}\,dx \quad \text{[M1]}$$ $$= \frac{x^4}{4}\ln x - \frac{1}{4}\int x^3\,dx = \frac{x^4}{4}\ln x - \frac{x^4}{16} + C = \frac{x^4}{16}(4\ln x - 1) + C \quad \text{[A1]}$$Evaluate $\int \frac{7x - 1}{x^2 - x - 6}\,dx$.
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Factor denominator: $x^2 - x - 6 = (x - 3)(x + 2)$.
$$\frac{7x - 1}{(x - 3)(x + 2)} = \frac{A}{x - 3} + \frac{B}{x + 2} \quad \text{[M1]}$$ $$7x - 1 = A(x + 2) + B(x - 3)$$• $x = 3$: $20 = 5A \implies A = 4$. [A1]
• $x = -2$: $-15 = -5B \implies B = 3$. [A1]
Evaluate $\int_0^1 x^2 e^{2x}\,dx$, giving your answer in the form $\frac{a e^2 + b}{c}$ where $a, b, c \in \mathbb{Z}$.
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First application: $u = x^2 \implies du = 2x\,dx$; $dv = e^{2x}dx \implies v = \frac{1}{2}e^{2x}$. [M1]
$$\int_0^1 x^2 e^{2x}\,dx = \left[ \frac{1}{2}x^2 e^{2x} \right]_0^1 - \int_0^1 x e^{2x}\,dx = \frac{1}{2}e^2 - \int_0^1 x e^{2x}\,dx \quad \text{[A1]}$$Second application for $\int_0^1 x e^{2x}\,dx$: $u_1 = x \implies du_1 = dx$; $dv_1 = e^{2x}dx \implies v_1 = \frac{1}{2}e^{2x}$. [M1]
$$\int_0^1 x e^{2x}\,dx = \left[ \frac{1}{2}x e^{2x} \right]_0^1 - \int_0^1 \frac{1}{2}e^{2x}\,dx = \frac{1}{2}e^2 - \left[ \frac{1}{4}e^{2x} \right]_0^1 = \frac{1}{2}e^2 - \left(\frac{1}{4}e^2 - \frac{1}{4}\right) = \frac{1}{4}e^2 + \frac{1}{4} \quad \text{[A1]}$$Subtract the second result from the boundary term: [M1]
$$\frac{1}{2}e^2 - \left(\frac{1}{4}e^2 + \frac{1}{4}\right) = \frac{1}{4}e^2 - \frac{1}{4} = \frac{e^2 - 1}{4} \quad \text{[A1]}$$Find $\int \arctan(2x)\,dx$.
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Insert factor $1$: $u = \arctan(2x) \implies du = \frac{2}{1 + (2x)^2}\,dx = \frac{2}{1 + 4x^2}\,dx$; $dv = 1\,dx \implies v = x$. [M1]
$$\int \arctan(2x)\,dx = x\arctan(2x) - \int \frac{2x}{1 + 4x^2}\,dx \quad \text{[M1]}$$Recognize derivative in numerator: $\frac{d}{dx}(1 + 4x^2) = 8x \implies \frac{2x}{1 + 4x^2} = \frac{1}{4}\frac{8x}{1 + 4x^2}$. [M1]
$$= x\arctan(2x) - \frac{1}{4}\ln(1 + 4x^2) + C \quad \text{[A1]}$$Find $\int \frac{4x - 1}{(x - 2)^2}\,dx$.
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Decompose with repeated linear factor template: [M1]
$$\frac{4x - 1}{(x - 2)^2} = \frac{A}{x - 2} + \frac{B}{(x - 2)^2} \implies 4x - 1 = A(x - 2) + B$$• Let $x = 2$: $4(2) - 1 = B \implies B = 7$. [A1]
• Compare $x$ coefficient: $A = 4$. [A1]
Examiner Pitfall: The second integral is a power rule $\int (x-2)^{-2}dx = -(x-2)^{-1}$, NOT a logarithm!
Evaluate $\int_0^1 \frac{3x^2 + 8x + 3}{x^2 + 3x + 2}\,dx$.
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Degrees are equal ($\text{deg } 2$). Must perform polynomial long division first: [M1]
$$\frac{3x^2 + 8x + 3}{x^2 + 3x + 2} = 3 + \frac{(8x + 3) - 3(3x + 2)}{x^2 + 3x + 2} = 3 + \frac{-x - 3}{(x + 1)(x + 2)} \quad \text{[A1]}$$Decompose remainder fraction: $\frac{-(x + 3)}{(x + 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x + 2}$ [M1]
$-(x + 3) = A(x + 2) + B(x + 1)$.
• $x = -1$: $-2 = A(1) \implies A = -2$.
• $x = -2$: $-1 = B(-1) \implies B = 1$.
Find the antiderivative $\int e^{2x}\sin(x)\,dx$.
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Let $I = \int e^{2x}\sin x\,dx$. Set $u = \sin x, dv = e^{2x}dx \implies du = \cos x\,dx, v = \frac{1}{2}e^{2x}$. [M1]
$$I = \frac{1}{2}e^{2x}\sin x - \frac{1}{2}\int e^{2x}\cos x\,dx \quad \text{[A1]}$$Apply parts to $\int e^{2x}\cos x\,dx$: $u_1 = \cos x, dv_1 = e^{2x}dx \implies du_1 = -\sin x\,dx, v_1 = \frac{1}{2}e^{2x}$. [M1]
$$\int e^{2x}\cos x\,dx = \frac{1}{2}e^{2x}\cos x + \frac{1}{2}\int e^{2x}\sin x\,dx = \frac{1}{2}e^{2x}\cos x + \frac{1}{2}I$$Substitute back into equation for $I$: [M1]
$$I = \frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x - \frac{1}{4}I \implies \frac{5}{4}I = \frac{e^{2x}(2\sin x - \cos x)}{4}$$ $$I = \frac{e^{2x}(2\sin x - \cos x)}{5} + C \quad \text{[A1]}$$Let $I_n = \int_1^e (\ln x)^n\,dx$ for $n \ge 1$.
(a) Prove that $I_n = e - n I_{n-1}$.
(b) Evaluate $I_0$ and hence find the exact value of $I_3$.
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(a) Set $u = (\ln x)^n, dv = 1\,dx \implies du = n(\ln x)^{n-1}\frac{1}{x}dx, v = x$. [M1]
$$I_n = \left[ x(\ln x)^n \right]_1^e - \int_1^e x \cdot n(\ln x)^{n-1}\frac{1}{x}\,dx = (e \cdot 1^n - 0) - n\int_1^e (\ln x)^{n-1}\,dx \quad \text{[M1]}$$ $$I_n = e - n I_{n-1} \quad \text{[AG]}$$(b) Base case and cascading evaluation:
$$I_0 = \int_1^e 1\,dx = [x]_1^e = e - 1 \quad \text{[A1]}$$ $$I_1 = e - 1(I_0) = e - (e - 1) = 1 \quad \text{[A1]}$$ $$I_2 = e - 2(I_1) = e - 2(1) = e - 2 \quad \text{[A1]}$$ $$I_3 = e - 3(I_2) = e - 3(e - 2) = e - 3e + 6 = 6 - 2e \quad \text{[A1]}$$Evaluate $\int \frac{2x^2 - x + 4}{x(x^2 + 4)}\,dx$.
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Set up decomposition with linear numerator over the quadratic factor: [M1]
$$\frac{2x^2 - x + 4}{x(x^2 + 4)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4} \implies 2x^2 - x + 4 = A(x^2 + 4) + (Bx + C)x$$• Let $x = 0$: $4 = 4A \implies A = 1$. [A1]
• Compare $x^2$: $2 = A + B \implies 2 = 1 + B \implies B = 1$. [A1]
• Compare $x$: $-1 = C \implies C = -1$. [A1]
Integrate each component separately: [M1]
$$\int \frac{1}{x}\,dx + \int \frac{x - 1}{x^2 + 4}\,dx = \ln|x| + \frac{1}{2}\int \frac{2x}{x^2 + 4}\,dx - \int \frac{1}{x^2 + 4}\,dx$$ $$= \ln|x| + \frac{1}{2}\ln(x^2 + 4) - \frac{1}{2}\arctan\left(\frac{x}{2}\right) + C \quad \text{[A1]}$$Examiner Tip: Notice how the linear term $\frac{x}{x^2+4}$ yields a logarithm, while the constant term $\frac{-1}{x^2+4}$ yields an $\arctan$. Don't conflate them!
IB Formula Booklet References (Topic 5.13 AHL)
Integration by Parts
$$\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx \quad \text{or} \quad \int u\,dv = uv - \int v\,du$$ $$\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du$$Standard Inverse Trigonometric and Rational Antiderivatives
$$\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$$ $$\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\left(\frac{x}{a}\right) + C \quad (|x| < a)$$ $$\int \frac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$$Key Formulas & Identities
1. The LIATE Priority Hierarchy
Logarithmic $\to$ Inverse Trigonometric $\to$ Algebraic $\to$ Trigonometric $\to$ Exponential
2. Partial Fraction Decomposition Templates
• Distinct linear: $\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$
• Repeated linear: $\frac{P(x)}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}$
• Irreducible quadratic: $\frac{P(x)}{(x-a)(x^2+c^2)} = \frac{A}{x-a} + \frac{Bx + D}{x^2 + c^2}$
• Improper fraction ($\text{deg } P \ge \text{deg } Q$): Divide first $\frac{P}{Q} = S(x) + \frac{R(x)}{Q(x)}$ before decomposing!
Common Exam Pitfalls
Sign Errors in Cyclic Integrals
When applying parts twice to $I = \int e^{ax}\cos(bx)\,dx$, students frequently lose track of negatives: $(-)\times(-) = (+)$, but subtract again when bringing to the left-hand side. Always write out parentheses explicitly around the second integral.
Missing $\frac{1}{a}$ Factor in Arctangent
Remember that $\int \frac{1}{x^2 + 16}\,dx = \frac{1}{4}\arctan\left(\frac{x}{4}\right) + C$. Dropping the external $\frac{1}{4}$ scalar is one of the most common errors in IB HL Paper 1.
Decomposing Improper Fractions Prematurely
Setting up $\frac{x^2}{x^2 - 4} = \frac{A}{x-2} + \frac{B}{x+2}$ leads to incorrect constants ($A=1, B=-1 \implies \frac{4}{x^2-4}$, losing the integer $1$). Always perform polynomial division when degree of numerator $\ge$ degree of denominator.
Topic Diagnostic Test
Complete this timed test to check your understanding and exam technique.
Section A: Diagnostic Multiple Choice Questions
According to the LIATE hierarchy, which function should be chosen as $u$ in $\int x^3 \ln x\,dx$?
What is $\int \frac{1}{x^2 + 16}\,dx$?
What is $\int x e^x\,dx$?
Why can $\int \ln x\,dx$ be solved by integration by parts?
Section B: Examination-Style Free Response Questions
Consider the integral $I = \int_1^e x^2 \ln x\,dx$.
(a) Use integration by parts to show that $I = \left[ \frac{x^3}{3}\ln x \right]_1^e - \int_1^e \frac{x^2}{3}\,dx$. [3 marks]
(b) Hence evaluate the exact value of $I$ in terms of $e$. [3 marks]
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(a) Parts set up: [3 marks]
$u = \ln x \implies du = \frac{1}{x}dx$ [M1].
$dv = x^2 dx \implies v = \frac{x^3}{3}$ [M1].
$I = \left[ \frac{x^3}{3}\ln x \right]_1^e - \int_1^e \frac{x^3}{3}\left(\frac{1}{x}\right)dx = \left[ \frac{x^3}{3}\ln x \right]_1^e - \int_1^e \frac{x^2}{3}\,dx$ [A1].
(b) Evaluation: [3 marks]
First term: $\left( \frac{e^3}{3}\ln e \right) - \left( \frac{1}{3}\ln 1 \right) = \frac{e^3}{3} - 0 = \frac{e^3}{3}$ [A1].
Second term: $\int_1^e \frac{x^2}{3}\,dx = \left[ \frac{x^3}{9} \right]_1^e = \frac{e^3 - 1}{9}$ [A1].
$I = \frac{e^3}{3} - \frac{e^3 - 1}{9} = \frac{3e^3 - e^3 + 1}{9} = \frac{2e^3 + 1}{9}$ [A1].
A curve has gradient $\frac{dy}{dx} = \frac{x + 7}{(x - 1)(x + 3)}$.
(a) Express $\frac{dy}{dx}$ in partial fractions. [3 marks]
(b) Given that the curve passes through $(2, 5)$, find the equation of the curve in the form $y = f(x)$. [4 marks]
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(a) Partial fractions: [3 marks]
$\frac{x + 7}{(x - 1)(x + 3)} = \frac{A}{x - 1} + \frac{B}{x + 3}$ [M1].
$x + 7 = A(x + 3) + B(x - 1)$.
• $x = 1$: $8 = 4A \implies A = 2$ [A1].
• $x = -3$: $4 = -4B \implies B = -1$ [A1].
$\frac{dy}{dx} = \frac{2}{x - 1} - \frac{1}{x + 3}$.
(b) Integration & Particular Solution: [4 marks]
$y = \int \left(\frac{2}{x - 1} - \frac{1}{x + 3}\right)dx = 2\ln|x - 1| - \ln|x + 3| + C$ [M1][A1].
Substitute $(2, 5)$:
$5 = 2\ln(1) - \ln(5) + C = 0 - \ln 5 + C \implies C = 5 + \ln 5$ [M1].
$y = 2\ln|x - 1| - \ln|x + 3| + 5 + \ln 5 = \ln\left( \frac{5(x - 1)^2}{|x + 3|} \right) + 5$ [A1].