Quick Recall
Quick check — pick the correct statement in each question (60 seconds) before we extend the method.
$\displaystyle\int \frac{1}{ax+b}\, \text{d}x = \frac{1}{a}\ln|ax+b| + C$. This works because $\dfrac{\text{d}}{\text{d}x}\big(\ln|ax+b|\big) = \dfrac{a}{ax+b}$, and the $a$ cancels the extra factor from the chain rule.
With $a=2$: $\displaystyle\int \frac{5}{2x+3}\, \text{d}x = 5 \cdot \frac{1}{2}\ln|2x+3| + C = \frac{5}{2}\ln|2x+3| + C$. The factor $\frac{1}{a}=\frac{1}{2}$ is essential.
$(2x+3)(2x+1) = 4x^2 + 8x + 3$, so the division gives $4x^2 + 8x = (2x+3)(2x+1) - 3$: quotient $2x+1$, remainder $-3$. Today's method uses exactly this to turn an improper fraction into a proper one.
The Wall
Now try this definite integral:
$$\int_{-1}^{3} \dfrac{4x^2 + 8x}{2x + 3}\, \text{d}x = \;?$$
The numerator $4x^2 + 8x$ has degree 2. The denominator $2x + 3$ has degree 1. Our Lesson 1 formula $\int \dfrac{1}{ax+b}\, \text{d}x = \dfrac{1}{a}\ln|ax+b| + C$ requires a constant numerator — but $4x^2 + 8x$ is not a constant. We hit a wall.
The Need
We need a way to reduce a rational function whose numerator has degree $\geq$ the denominator, into a form we can integrate. The technique is polynomial long division followed by the $\ln$ formula from Lesson 1. Today we learn this two-step process.
💡 The Bridge (click to reveal)
Divide $4x^2 + 8x$ by $2x + 3$:
$$4x^2 + 8x = (2x + 3)(2x + 1) - 3$$
So:
$$\frac{4x^2 + 8x}{2x + 3} = 2x + 1 - \frac{3}{2x + 3}$$
Now each piece can be integrated:
$$\int \!\left(2x + 1 - \frac{3}{2x+3}\right) \text{d}x = x^2 + x - \frac{3}{2}\ln|2x+3| + C$$
The wall is climbed. The “ladder” is polynomial division.
Last lesson we learned $\int \dfrac{1}{ax+b}\, \text{d}x = \dfrac{1}{a}\ln|ax+b| + C$. But this only works when the numerator is a constant. Today we tackle rational functions where the numerator is a polynomial of degree $\geq 1$.
When $\deg P(x) \geq \deg(ax+b)$, we can write:
$$\boxed{\;\dfrac{P(x)}{ax+b} = Q(x) + \dfrac{R}{ax+b}\;}$$where $Q(x)$ is the quotient polynomial and $R$ is the (constant) remainder. The fraction is now proper (constant over linear) and can be integrated with the $\ln$ formula.
A fraction with a linear numerator and linear denominator can be rewritten by extracting the coefficient that matches the denominator's derivative:
$$\boxed{\;\dfrac{Ax+B}{ax+b} = \dfrac{A}{a} + \dfrac{B - \dfrac{Ab}{a}}{ax+b}\;}$$Equivalently, write $Ax + B = \dfrac{A}{a}(ax + b) + \left(B - \dfrac{Ab}{a}\right)$, then divide through by $ax+b$.
Example: $\dfrac{6x}{3x-2} = \dfrac{6}{3} + \dfrac{0 - \dfrac{6 \cdot (-2)}{3}}{3x-2} = 2 + \dfrac{4}{3x-2}$.
After division, the proper fraction $\dfrac{R}{ax+b}$ is integrated using the Lesson 1 formula:
$$\boxed{\;\int \dfrac{R}{ax+b}\, \text{d}x = \dfrac{R}{a}\ln|ax+b| + C\;}\qquad(R \text{ constant},\ a \neq 0)$$Combined with $\int Q(x)\, \text{d}x$ (power rule), this gives the full result.
To divide $4x^2 + 8x$ by $2x + 3$, we set up the long-division bracket (the "厂" shape — divisor on the left, dividend under the top bar):
In groups of three, drag each integral — or tap it and then tap a box — into the box that names the first step it needs. A card placed in the wrong box flashes red and returns to the pile, so read the degree of the numerator before you move. Six integrals, three preparation routes.
Group rule: the member holding the card must name the degree test and explain the route out loud before the group taps a box — nobody moves a card until all three members agree.
Two worked integrations by decomposition, each with exactly one wrong line. Work on your own: click the line you believe is wrong, then choose the corrected line from the options. One wrong line per round — reason it out, do not guess.
(i) Differentiate $\ln(2x + 3)$.
(ii) Hence, or otherwise, show that $\displaystyle\int_{-1}^{3} \frac{1}{2x + 3} \, \text{d}x = \ln 3$.
(iii) Find the quotient and remainder when $4x^2 + 8x$ is divided by $2x + 3$.
(iv) Hence show that $\displaystyle\int_{-1}^{3} \frac{4x^2 + 8x}{2x + 3} \, \text{d}x = 12 - 3\ln 3$.
(Cambridge International AS & A Level Mathematics 9709 Paper 2 Q7 June 2006)
👁 Reveal Solution
By the chain rule: $\dfrac{\text{d}}{\text{d}x}\!\left[\ln(2x+3)\right] = \dfrac{1}{2x+3} \cdot 2 = \dfrac{2}{2x+3}$.
$\boxed{\dfrac{\text{d}}{\text{d}x}\!\left[\ln(2x+3)\right] = \dfrac{2}{2x+3}}$
From (i), $\dfrac{\text{d}}{\text{d}x}\!\left[\ln(2x+3)\right] = \dfrac{2}{2x+3}$, so $\dfrac{\text{d}}{\text{d}x}\!\left[\dfrac{1}{2}\ln(2x+3)\right] = \dfrac{1}{2x+3}$.
Therefore $\displaystyle\int \dfrac{1}{2x+3}\, \text{d}x = \dfrac{1}{2}\ln|2x+3| + C$.
Evaluate: $\left[\dfrac{1}{2}\ln|2x+3|\right]_{-1}^{3} = \dfrac{1}{2}\ln 9 - \dfrac{1}{2}\ln 1 = \dfrac{1}{2}\ln 9 - 0$.
Since $\ln 9 = \ln 3^2 = 2\ln 3$, we have $\dfrac{1}{2} \cdot 2\ln 3 = \ln 3$.
$\boxed{\displaystyle\int_{-1}^{3} \dfrac{1}{2x+3}\, \text{d}x = \ln 3}$ ✓
Polynomial long division (vertical layout shown in the Key Concepts card above):
Step 1: $\dfrac{4x^2}{2x} = 2x$. Multiply: $2x(2x+3) = 4x^2 + 6x$. Subtract: $(4x^2+8x) - (4x^2+6x) = 2x$.
Step 2: $\dfrac{2x}{2x} = 1$. Multiply: $1(2x+3) = 2x + 3$. Subtract: $2x - (2x+3) = -3$.
So $4x^2 + 8x = (2x + 3)(2x + 1) - 3$.
$\boxed{\text{Quotient} = 2x + 1,\quad \text{Remainder} = -3}$ ✓
From (iii): $\dfrac{4x^2 + 8x}{2x + 3} = 2x + 1 - \dfrac{3}{2x + 3}$.
Split the integral:
$\displaystyle\int_{-1}^{3} \!\left(2x + 1 - \dfrac{3}{2x + 3}\right) \text{d}x = \int_{-1}^{3} (2x+1)\, \text{d}x - 3 \int_{-1}^{3} \dfrac{1}{2x+3}\, \text{d}x$
First piece: $\left[x^2 + x\right]_{-1}^{3} = (9 + 3) - (1 - 1) = 12 - 0 = 12$.
Second piece: by part (ii), $\displaystyle\int_{-1}^{3} \dfrac{1}{2x+3}\, \text{d}x = \ln 3$. So $-3 \cdot \ln 3 = -3\ln 3$.
Combine: $12 - 3\ln 3$.
$\boxed{\displaystyle\int_{-1}^{3} \dfrac{4x^2 + 8x}{2x + 3}\, \text{d}x = 12 - 3\ln 3}$ ✓
Verification (numerical): $12 - 3\ln 3 \approx 12 - 3(1.0986) = 12 - 3.2958 = 8.7042$.
(i) Find the equation of the tangent to the curve $y = \ln(3x - 2)$ at the point where $x = 1$.
(ii)(a) Find the value of the constant $A$ such that $\dfrac{6x}{3x - 2} \equiv 2 + \dfrac{A}{3x - 2}$.
(ii)(b) Hence show that $\displaystyle\int_{2}^{6} \frac{6x}{3x - 2} \, \text{d}x = 8 + \frac{8}{3} \ln 2$.
(Cambridge International AS & A Level Mathematics 9709 Paper 2 Q8 June 2009)
👁 Reveal Solution
Find $y(1)$: $y(1) = \ln(3 \cdot 1 - 2) = \ln 1 = 0$. So the point is $(1, 0)$.
Find $y'(1)$: $y' = \dfrac{3}{3x-2}$ (chain rule). So $y'(1) = \dfrac{3}{3-2} = 3$.
Point-slope form: $y - 0 = 3(x - 1)$, i.e. $y = 3x - 3$.
$\boxed{y = 3x - 3}$ ✓
Multiply both sides by $3x - 2$:
$6x = 2(3x - 2) + A = 6x - 4 + A$.
Equating constants: $0 = -4 + A$, so $A = 4$.
$\boxed{A = 4}$ ✓
So $\dfrac{6x}{3x-2} = 2 + \dfrac{4}{3x-2}$. (This is the “algebraic short-cut”: write $6x = 2(3x-2) + 4$.)
Using part (ii)(a): $\dfrac{6x}{3x-2} = 2 + \dfrac{4}{3x-2}$. Split the integral:
$\displaystyle\int_{2}^{6} \!\left(2 + \dfrac{4}{3x-2}\right) \text{d}x = \int_{2}^{6} 2\, \text{d}x + 4\int_{2}^{6} \dfrac{1}{3x-2}\, \text{d}x$
First piece: $\left[2x\right]_{2}^{6} = 12 - 4 = 8$.
Second piece: $\displaystyle\int \dfrac{1}{3x-2}\, \text{d}x = \dfrac{1}{3}\ln|3x-2|$. So $4 \cdot \left[\dfrac{1}{3}\ln|3x-2|\right]_{2}^{6} = \dfrac{4}{3}\left(\ln 16 - \ln 4\right) = \dfrac{4}{3}\ln\!\left(\dfrac{16}{4}\right) = \dfrac{4}{3}\ln 4$.
Since $\ln 4 = 2\ln 2$: $\dfrac{4}{3} \cdot 2\ln 2 = \dfrac{8}{3}\ln 2$.
Combine: $8 + \dfrac{8}{3}\ln 2$.
$\boxed{\displaystyle\int_{2}^{6} \dfrac{6x}{3x - 2}\, \text{d}x = 8 + \dfrac{8}{3}\ln 2}$ ✓
Verification (numerical): $8 + \dfrac{8}{3}\ln 2 \approx 8 + \dfrac{8}{3}(0.6931) = 8 + 1.8484 = 9.8484$.
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Find the exact value of $\displaystyle\int_{-1}^{6} \frac{1}{2x+3} \, \text{d}x$.
💡 Hint
Use the formula $\int \dfrac{1}{ax+b}\, \text{d}x = \dfrac{1}{a}\ln|ax+b|$ with $a = 2$, $b = 3$.
Evaluate at $-1$ and $6$, then simplify using $\ln$ laws.
👁 Reveal Solution
Here $a = 2$, $b = 3$. So $\displaystyle\int \dfrac{1}{2x+3}\, \text{d}x = \dfrac{1}{2}\ln|2x+3|$.
$\left[\dfrac{1}{2}\ln|2x+3|\right]_{-1}^{6} = \dfrac{1}{2}\ln|2(6)+3| - \dfrac{1}{2}\ln|2(-1)+3|$
$= \dfrac{1}{2}\ln 15 - \dfrac{1}{2}\ln 1 = \dfrac{1}{2}\ln 15 - 0$
$\boxed{\dfrac{1}{2}\ln 15}$ ✓
Numerical check: $\dfrac{1}{2}\ln 15 \approx \dfrac{1}{2}(2.7081) = 1.3540$.
Show that $\displaystyle\int_{6}^{16} \frac{6}{2x - 7} \, \text{d}x = \ln 125$.
[Adapted: part (i) only; the original also asks for a trapezium-rule approximation which is out of scope for this lesson.]
💡 Hint
Use $\int \dfrac{A}{ax+b}\, \text{d}x = \dfrac{A}{a}\ln|ax+b|$ with $A = 6$, $a = 2$, $b = -7$.
So the antiderivative is $\dfrac{6}{2}\ln|2x-7| = 3\ln|2x-7|$.
Evaluate at $6$ and $16$, then simplify using $\ln$ laws: $\ln a - \ln b = \ln\dfrac{a}{b}$ and $n \ln a = \ln a^n$.
👁 Reveal Solution
Here $A = 6$, $a = 2$, $b = -7$. So:
$\displaystyle\int \dfrac{6}{2x - 7}\, \text{d}x = \dfrac{6}{2}\ln|2x-7| = 3\ln|2x-7|$
$\left[3\ln|2x-7|\right]_{6}^{16} = 3\ln|2(16)-7| - 3\ln|2(6)-7|$
$= 3\ln 25 - 3\ln 5$
$= 3(\ln 25 - \ln 5) = 3\ln\!\left(\dfrac{25}{5}\right) = 3\ln 5$
Using $n \ln a = \ln a^n$: $3\ln 5 = \ln 5^3 = \ln 125$.
$\boxed{\displaystyle\int_{6}^{16} \dfrac{6}{2x - 7}\, \text{d}x = \ln 125}$ ✓
Numerical check: $\ln 125 \approx 4.8283$. Direct computation: $3\ln 5 \approx 3(1.6094) = 4.8283$. ✓
The diagram illustrates the graph of $y = e^x$. The point A has coordinates $(\ln 5, 0)$, B has coordinates $(\ln 5, 5)$ and C has coordinates $(0, 5)$.
(i) Find the area of the region OABE enclosed by the curve $y = e^x$, the $x$-axis, the $y$-axis and the line AB. Hence find the area of the shaded region EBC.
(ii) The graph of $y = e^x$ is transformed into the graph of $y = \ln x$. Describe this transformation geometrically.
(iii) Using your answers to parts (i) and (ii), or otherwise, show that $\displaystyle\int_1^5 \ln x \, \text{d}x = 5\ln 5 - 4$.
(iv) Deduce the values of (a) $\displaystyle\int_1^5 \ln(x^3) \, \text{d}x$ (b) $\displaystyle\int_1^5 \ln(3x) \, \text{d}x$
💡 Hint
For (i): the area OABE is $\int_0^{\ln 5} e^x\, \text{d}x$. The rectangle OABC has area $5 \cdot \ln 5$. The shaded EBC = rectangle − OABE.
For (ii): $y = e^x$ and $y = \ln x$ are inverse functions. The graph of an inverse function is the reflection of the original in the line $y = x$.
For (iii): the area under $\ln x$ from 1 to 5 equals the shaded EBC region (by symmetry of inverse functions).
For (iv)(a): use $\ln(x^3) = 3\ln x$. For (iv)(b): use $\ln(3x) = \ln 3 + \ln x$.
👁 Reveal Solution
OABE = $\displaystyle\int_0^{\ln 5} e^x\, \text{d}x = \left[e^x\right]_0^{\ln 5} = e^{\ln 5} - e^0 = 5 - 1 = 4$.
Rectangle OABC = width $\ln 5$ × height $5 = 5\ln 5$.
Shaded EBC = rectangle − OABE = $5\ln 5 - 4$.
$\boxed{\text{OABE} = 4, \quad \text{EBC} = 5\ln 5 - 4}$ ✓
$y = e^x$ and $y = \ln x$ are inverse functions.
$\boxed{\text{Reflection in the line } y = x}$ ✓
By the reflection in $y = x$, the area under $\ln x$ from $1$ to $5$ equals the shaded region EBC found in part (i).
(The region under $\ln x$ for $1 \leq x \leq 5$ is the mirror image of EBC.)
So $\displaystyle\int_1^5 \ln x\, \text{d}x = 5\ln 5 - 4$.
$\boxed{\displaystyle\int_1^5 \ln x\, \text{d}x = 5\ln 5 - 4}$ ✓
Using $\ln(x^3) = 3\ln x$:
$\displaystyle\int_1^5 \ln(x^3)\, \text{d}x = 3\int_1^5 \ln x\, \text{d}x = 3(5\ln 5 - 4) = 15\ln 5 - 12$.
$\boxed{15\ln 5 - 12}$ ✓
Using $\ln(3x) = \ln 3 + \ln x$:
$\displaystyle\int_1^5 \ln(3x)\, \text{d}x = \int_1^5 \ln 3\, \text{d}x + \int_1^5 \ln x\, \text{d}x$
$= \ln 3 \cdot (5 - 1) + (5\ln 5 - 4) = 4\ln 3 + 5\ln 5 - 4$.
$\boxed{5\ln 5 + 4\ln 3 - 4}$ ✓
The graph of $y = x + \dfrac{4}{x}$ is shown below.
(i) Find the coordinates of the minimum point, P, and the maximum point, Q.
(ii) Find the area of each shaded region.
[Note: the original exercise includes a figure showing two shaded regions — one in the first quadrant between the curve, the $x$-axis, $x = 1$, and $x = 2$; the other in the third quadrant between the curve, the $x$-axis, $x = -2$, and $x = -1$. Refer to the Hodder textbook page 140 for the figure.]
💡 Hint
For (i): set $\dfrac{\text{d}y}{\text{d}x} = 1 - \dfrac{4}{x^2} = 0$ and solve. Then use the second derivative $\dfrac{\text{d}^2y}{\text{d}x^2} = \dfrac{8}{x^3}$ to classify.
For (ii): the curve $y = x + \dfrac{4}{x}$ is above the $x$-axis on $[1, 2]$ (in Q1) and below on $[-2, -1]$ (in Q3). Compute $\int_1^2 \!\left(x + \dfrac{4}{x}\right) \text{d}x$ and $\left|\int_{-2}^{-1} \!\left(x + \dfrac{4}{x}\right) \text{d}x\right|$.
The integral $\int \dfrac{4}{x}\, \text{d}x = 4\ln|x|$ — this is the ln-formula with $A = 4$, $a = 1$, $b = 0$.
👁 Reveal Solution
$y = x + \dfrac{4}{x}$. Differentiate: $\dfrac{\text{d}y}{\text{d}x} = 1 - \dfrac{4}{x^2}$.
Set $\dfrac{\text{d}y}{\text{d}x} = 0$: $1 - \dfrac{4}{x^2} = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$.
Second derivative: $\dfrac{\text{d}^2y}{\text{d}x^2} = \dfrac{8}{x^3}$.
At $x = 2$: $\dfrac{\text{d}^2y}{\text{d}x^2} = \dfrac{8}{8} = 1 > 0$, so minimum. $y(2) = 2 + 2 = 4$. So $P = (2, 4)$.
At $x = -2$: $\dfrac{\text{d}^2y}{\text{d}x^2} = \dfrac{8}{-8} = -1 < 0$, so maximum. $y(-2) = -2 - 2 = -4$. So $Q = (-2, -4)$.
$\boxed{P = (2, 4) \text{ (minimum)}, \quad Q = (-2, -4) \text{ (maximum)}}$ ✓
First find the antiderivative: $\displaystyle\int \!\left(x + \dfrac{4}{x}\right) \text{d}x = \dfrac{x^2}{2} + 4\ln|x| + C$.
Region 1 (Q1, between $x = 1$ and $x = 2$): the curve is above the $x$-axis here.
$\displaystyle\int_1^2 \!\left(x + \dfrac{4}{x}\right) \text{d}x = \left[\dfrac{x^2}{2} + 4\ln|x|\right]_1^2 = \left(\dfrac{4}{2} + 4\ln 2\right) - \left(\dfrac{1}{2} + 4\ln 1\right)$
$= (2 + 4\ln 2) - \left(\dfrac{1}{2} + 0\right) = \dfrac{3}{2} + 4\ln 2$.
Region 2 (Q3, between $x = -2$ and $x = -1$): the curve is below the $x$-axis, so area = $-\displaystyle\int_{-2}^{-1} \!\left(x + \dfrac{4}{x}\right) \text{d}x$.
$\displaystyle\int_{-2}^{-1} \!\left(x + \dfrac{4}{x}\right) \text{d}x = \left[\dfrac{x^2}{2} + 4\ln|x|\right]_{-2}^{-1} = \left(\dfrac{1}{2} + 4\ln 1\right) - \left(\dfrac{4}{2} + 4\ln 2\right)$
$= \dfrac{1}{2} - 2 - 4\ln 2 = -\dfrac{3}{2} - 4\ln 2$.
Area = $-\left(-\dfrac{3}{2} - 4\ln 2\right) = \dfrac{3}{2} + 4\ln 2$.
$\boxed{\text{Region 1} = \dfrac{3}{2} + 4\ln 2, \quad \text{Region 2} = \dfrac{3}{2} + 4\ln 2}$ ✓
Both regions have the same area by the odd-function symmetry of $y = x + \dfrac{4}{x}$. Numerical check: $\dfrac{3}{2} + 4\ln 2 \approx 1.5 + 4(0.6931) = 1.5 + 2.7726 = 4.2726$.
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.