7.2 Integrals of Some Particular Functions and Partial Fractions
Six special integrals handle $\dfrac{1}{x^2\pm a^2}$, $\dfrac{1}{a^2-x^2}$ and their square roots. Complete the square and every quadratic denominator becomes one of them; split a rational function into partial fractions and it falls apart into logarithms and inverse tangents.
A raindrop falling through air speeds up, but the drag grows with the square of its speed, so its acceleration is something like $9.8-0.2v^2$. How long does it take to reach $5$ m/s? The answer needs $\displaystyle\int\frac{dv}{49-v^2}$. A chemist timing a reaction between two substances meets $\displaystyle\int\frac{dx}{(3-x)(5-x)}$; an ecologist modelling a fish population that levels off meets $\displaystyle\int\frac{dP}{P(100-P)}$. None of these is in the table of Section 7.1, and no simple substitution removes them. Yet they share a shape: one over a quadratic, or more generally a ratio of polynomials.
This section builds the two tools that handle that shape. First, six special integrals — each checked by differentiation once and then used as a formula — cover $\dfrac{1}{x^2\pm a^2}$, $\dfrac{1}{a^2-x^2}$ and their square-root versions; completing the square turns any quadratic into one of them. Second, partial fractions break a rational function into a sum of simple fractions, each of which integrates to a logarithm, a power or an inverse tangent.
Six Special Integrals
Throughout, $a$ is a positive constant. The first three integrands have a quadratic denominator and no root.
📐 Theorem 7.6 — Special integrals with $x^2\pm a^2$ and $a^2-x^2$
For $a\gt0$, on any interval where the integrand is defined:
Why. For (1), split the fraction: since $(x+a)-(x-a)=2a$,
$$\frac{1}{x^2-a^2}=\frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right)$$
Integrate each part to get $\dfrac{1}{2a}\log|x-a|$ $-\dfrac{1}{2a}\log|x+a|$, which is the formula. Formula (2) is proved the same way from $\dfrac{1}{a^2-x^2}=\dfrac{1}{2a}\left(\dfrac{1}{a-x}+\dfrac{1}{a+x}\right)$. For (3), substitute $x=a\tan\theta$, $dx=a\sec^2\theta\,d\theta$: the integrand becomes $\dfrac{a\sec^2\theta}{a^2\sec^2\theta}=\dfrac1a$, so the integral is $\dfrac{\theta}{a}+C=\dfrac1a\tan^{-1}\dfrac xa+C$.
The next three have a square root in the denominator.
📐 Theorem 7.7 — Special integrals with square roots
For $a\gt0$, on any interval where the integrand is defined:
(4) root of x² − a²$\displaystyle\int\frac{dx}{\sqrt{x^2-a^2}}=\log\left|x+\sqrt{x^2-a^2}\right|+C$, $\ |x|\gt a$
(5) root of a² − x²$\displaystyle\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac xa+C$, $\ |x|\lt a$
(6) root of x² + a²$\displaystyle\int\frac{dx}{\sqrt{x^2+a^2}}=\log\left|x+\sqrt{x^2+a^2}\right|+C$
Why. For (5), put $x=a\sin\theta$ with $\theta\in\left(-\tfrac\pi2,\tfrac\pi2\right)$; then $\sqrt{a^2-x^2}=a\cos\theta$ and $dx=a\cos\theta\,d\theta$, so the integral is $\displaystyle\int d\theta=\theta+C$. For (6), differentiate the answer. With $s=\sqrt{x^2+a^2}$, $\dfrac{ds}{dx}=\dfrac xs$, so
$$\frac{d}{dx}\log(x+s)=\frac{1+\frac xs}{x+s}=\frac{\frac{s+x}{s}}{x+s}=\frac1s$$
Formula (4) is checked in exactly the same way with $s=\sqrt{x^2-a^2}$. Notice the pattern: plus $a^2$ under the root or in the denominator without a root gives an inverse tangent or a logarithm of $x+\sqrt{\ }$; minus gives logarithms of a ratio, or an inverse sine.
(a) Formula (1) with $a=4$: $\dfrac18\log\left|\dfrac{x-4}{x+4}\right|+C$.
(b) The coefficient of $x^2$ must be $1$. Put $u=2x$, $dx=\tfrac12du$:
$$\frac12\int\frac{du}{\sqrt{9-u^2}}=\frac12\sin^{-1}\frac u3+C=\frac12\sin^{-1}\frac{2x}{3}+C$$
(c) Take out the $4$: $\displaystyle\frac14\int\frac{dx}{x^2+\frac94}$, then formula (3) with $a=\tfrac32$:
$$\frac14\cdot\frac23\tan^{-1}\frac{2x}{3}+C=\frac16\tan^{-1}\frac{2x}{3}+C$$
(d) $x^4=(x^2)^2$ and $x\,dx$ is half of $d(x^2)$. With $u=x^2$: $\displaystyle\frac12\int\frac{du}{\sqrt{u^2+1}}=\frac12\log\left|x^2+\sqrt{x^4+1}\right|+C$.
Check (c): $\dfrac{d}{dx}\left(\tfrac16\tan^{-1}\tfrac{2x}3\right)=\dfrac16\cdot\dfrac{2/3}{1+\frac{4x^2}{9}}$, which simplifies to $\dfrac{1}{9+4x^2}$ ✓.
Quadratic Denominators: Completing the Square
A quadratic $ax^2+bx+c$ in a denominator, with or without a square root, is brought to one of the six forms by completing the square:
$$ax^2+bx+c=a\left[\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a^2}\right]$$
With $t=x+\dfrac{b}{2a}$ (so $dt=dx$), the bracket is $t^2+k^2$ if $4ac-b^2\gt0$, or $t^2-k^2$ if $4ac-b^2\lt0$. The sign of the discriminant $b^2-4ac$ decides which special integral you will need: no real roots gives an inverse tangent; two real roots gives a logarithm.
(a) $x^2+4x+13=(x+2)^2+9$. Formula (3) with $t=x+2$, $a=3$: $\ \dfrac13\tan^{-1}\dfrac{x+2}{3}+C$.
(b) square Take out the $2$: $2x^2+x-3=2\left[\left(x+\tfrac14\right)^2-\tfrac{25}{16}\right]$, a difference of squares with $a=\tfrac54$.
(b) formula By (1),
$$\frac12\cdot\frac{1}{2\cdot\frac54}\log\left|\frac{x+\frac14-\frac54}{x+\frac14+\frac54}\right|+C=\frac15\log\left|\frac{2x-2}{2x+3}\right|+C$$
(multiply top and bottom of the fraction by $2$; equivalently $\tfrac15\log\left|\tfrac{x-1}{2x+3}\right|$ up to a constant).
(c) $5-4x-x^2=9-(x^2+4x+4)$ $=9-(x+2)^2$. Formula (5): $\ \sin^{-1}\dfrac{x+2}{3}+C$, valid for $-5\lt x\lt1$.
Watch In (b) and (c) a negative leading coefficient or a factor in front must be handled before completing the square; the factor $\tfrac12$ in (b) is easy to lose.
Interactive 1 — Which Special Integral? Let the Discriminant Decide
Left: the parabola $q(x)=x^2+bx+c$. Right: the integrand $\dfrac{1}{q(x)}$. Completing the square moves the vertex to $x=-\tfrac b2$ and leaves $t^2+k$ with $t=x+\tfrac b2$, where $k$ is the height of the vertex. If the vertex is above the axis ($k\gt0$), $q$ never vanishes and the answer is an inverse tangent. If it is below ($k\lt0$), $q$ has two roots — dashed asymptotes of $\tfrac1q$ — and the answer is a logarithm. Play slides $c$ through the boundary case $k=0$.
$q(x)=x^2+bx+c$ and its vertex
the integrand $1/q(x)$
completed square—
form—
$\int\frac{dx}{q(x)}$—
When the numerator is a linear expression $px+q$, split it into a multiple of the derivative of the quadratic plus a constant. The first piece then integrates by the $\dfrac{f'}{f}$ rule (or to a square root), the second by a special integral.
🧭 Problem-solving strategy — a linear numerator over a quadratic (or its square root)
For $\displaystyle\int\frac{px+q}{ax^2+bx+c}\,dx$ and $\displaystyle\int\frac{px+q}{\sqrt{ax^2+bx+c}}\,dx$:
Write $px+q=A\,\dfrac{d}{dx}\left(ax^2+bx+c\right)+B$, that is, $px+q=A(2ax+b)+B$. Compare coefficients: $2aA=p$ fixes $A$, then $Ab+B=q$ fixes $B$.
The $A$-part: $\displaystyle A\int\frac{2ax+b}{ax^2+bx+c}\,dx=A\log\left|ax^2+bx+c\right|$, or, with a square root, $\displaystyle A\int\frac{2ax+b}{\sqrt{ax^2+bx+c}}\,dx=2A\sqrt{ax^2+bx+c}$ (substitute $u=ax^2+bx+c$).
The $B$-part: $\displaystyle B\int\frac{dx}{ax^2+bx+c}$ (or with the root) — complete the square and use one of the six special integrals.
Add the two parts and one constant $C$.
✏️ Example 3 — A linear numerator over a quadratic
Find $\displaystyle\int\frac{x+3}{x^2-2x+5}\,dx$.
💡 Show solution▶
Split $\dfrac{d}{dx}(x^2-2x+5)=2x-2$. Write $x+3=A(2x-2)+B$: $2A=1$ gives $A=\tfrac12$, and $-2A+B=3$ gives $B=4$. So $x+3=\tfrac12(2x-2)+4$.
A-part $\displaystyle\frac12\int\frac{2x-2}{x^2-2x+5}\,dx=\frac12\log\left(x^2-2x+5\right)$ — no modulus needed, as $x^2-2x+5=(x-1)^2+4\gt0$.
You know how to add fractions: $\dfrac{2}{x-1}+\dfrac{1}{x+3}=\dfrac{3x+5}{(x-1)(x+3)}$. To integrate the right-hand side, we need to go backwards — to split one complicated fraction into simple ones whose integrals we know. That reverse process is called splitting into partial fractions.
📘 Definition 7.3 — Proper and improper rational functions
A rational function is a quotient $\dfrac{P(x)}{Q(x)}$ of polynomials with $Q(x)\ne0$. It is a proper rational function if the degree of $P$ is less than the degree of $Q$, and an improper rational function otherwise. Long division writes every improper rational function as a polynomial plus a proper one:
$$\frac{P(x)}{Q(x)}=T(x)+\frac{R(x)}{Q(x)},\qquad \deg R\lt\deg Q$$
📘 Definition 7.4 — Partial fraction decomposition
Writing a proper rational function as a sum of simpler fractions, one for each factor of the denominator, is called its partial fraction decomposition. Here $a,b,c$ are distinct real numbers, and $x^2+k$ with $k\gt0$ has no real factors (an irreducible quadratic factor):
distinct linear factors$\dfrac{px+q}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}$
three distinct linear factors$\dfrac{px^2+qx+r}{(x-a)(x-b)(x-c)}$ $=\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{x-c}$
a repeated linear factor$\dfrac{px+q}{(x-a)^2}=\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}$
repeated factor and another factor$\dfrac{px^2+qx+r}{(x-a)^2(x-b)}$ $=\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}+\dfrac{C}{x-b}$
linear times irreducible quadratic$\dfrac{px^2+qx+r}{(x-a)(x^2+k)}$ $=\dfrac{A}{x-a}+\dfrac{Bx+C}{x^2+k}$
The constants $A,B,C$ are found by clearing denominators and then either substituting convenient values of $x$ or comparing coefficients.
A fast way to find the constants. For a non-repeated linear factor $x-a$, multiply the identity by $x-a$ and put $x=a$: every other fraction is killed, and $A$ is the rest of the original fraction evaluated at $x=a$ — in effect, cover up the factor $x-a$ and substitute. This cover-up rule finds every constant belonging to a non-repeated linear factor. Constants for repeated or quadratic factors are then found by comparing coefficients, or by substituting one or two more values of $x$.
✏️ Example 5 — Distinct linear factors, and an improper fraction
(a) constants $\dfrac{3x+5}{(x-1)(x+3)}=\dfrac{A}{x-1}+\dfrac{B}{x+3}$. Cover up $x-1$ and put $x=1$: $A=\dfrac{8}{4}=2$. Cover up $x+3$ and put $x=-3$: $B=\dfrac{-4}{-4}=1$.
(b) divide The fraction is improper (degree $3$ over degree $2$). Division: $x^3+x+1=x\left(x^2-1\right)+(2x+1)$, so the integrand is $x+\dfrac{2x+1}{(x-1)(x+1)}$.
(b) split Cover-up: at $x=1$, $A=\tfrac32$; at $x=-1$, $B=\dfrac{-1}{-2}=\tfrac12$. Hence
$$\int\frac{x^3+x+1}{x^2-1}\,dx=\frac{x^2}{2}+\frac32\log|x-1|+\frac12\log|x+1|+C$$
(a) form $\dfrac{3x-2}{(x+1)^2(x+3)}=\dfrac{A}{x+1}+\dfrac{B}{(x+1)^2}+\dfrac{C}{x+3}$, that is, $3x-2=A(x+1)(x+3)$ $+B(x+3)+C(x+1)^2$.
(a) constants $x=-1$: $-5=2B$, $B=-\tfrac52$. $\ x=-3$: $-11=4C$, $C=-\tfrac{11}4$. Coefficient of $x^2$: $0=A+C$, so $A=\tfrac{11}4$. (Check at $x=0$: $\tfrac{11}4\cdot3-\tfrac52\cdot3-\tfrac{11}4=-2$ ✓.)
(a) integrate $\displaystyle\int\frac{B}{(x+1)^2}dx=-\frac{B}{x+1}$, so
$$\frac{11}{4}\log|x+1|+\frac{5}{2(x+1)}-\frac{11}{4}\log|x+3|+C$$
(b) constants $\dfrac{1}{(x+1)(x^2+1)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+1}$, so $1=A(x^2+1)+(Bx+C)(x+1)$. $x=-1$: $A=\tfrac12$. Coefficient of $x^2$: $0=A+B$, $B=-\tfrac12$. Constant term: $1=A+C$, $C=\tfrac12$.
(b) integrate $\displaystyle\int\frac{-\frac12x+\frac12}{x^2+1}dx=-\frac14\log(x^2+1)+\frac12\tan^{-1}x$, so
$$\int\frac{dx}{(x+1)(x^2+1)}=\frac12\log|x+1|-\frac14\log\left(x^2+1\right)+\frac12\tan^{-1}x+C$$
Interactive 2 — Partial-Fractions Decomposer with a Recombination Check
Choose a type and set the numerator $px+q$ and the roots. The lab finds the constants (cover-up for each non-repeated factor), draws the original fraction in black and the two partial fractions in blue and green. Their sum, dashed red, lies exactly on the black curve — that is the recombination check. Play moves a probe along $x$ and adds the two pieces numerically.
fraction—
partial fractions—
probe: sum of pieces—
⚠️ Common mistakes when studying alone
1. Splitting an improper fraction. Partial fractions are only for proper fractions. If the degree of the numerator is not smaller, divide first (Example 5(b)).
2. Missing a term for a repeated factor. $(x+1)^2$ needs both $\dfrac{A}{x+1}$ and $\dfrac{B}{(x+1)^2}$. With only one of them, the constants cannot be found.
3. A constant over a quadratic factor. Over $x^2+1$ the numerator is $Bx+C$, not just $B$.
4. Using the wrong special integral. $\dfrac{1}{x^2+4}$ gives $\tfrac12\tan^{-1}\tfrac x2$, but $\dfrac{1}{x^2-4}$ gives a logarithm. Look at the sign before reaching for a formula.
5. Dropping the factor in front. In $\dfrac{1}{4x^2+9}$ or $\dfrac{1}{2x^2+x-3}$ take out the leading coefficient first; forgetting it changes the answer by a constant factor, which no $+\,C$ can repair.
Exercises
I
The six special integrals. Reduce each to one of formulas (1)–(6).
1 Formula (3), $a=5$: $\tfrac15\tan^{-1}\dfrac x5+C$.
2 Formula (1), $a=3$: $\tfrac16\log\left|\dfrac{x-3}{x+3}\right|+C$.
3 Formula (5), $a=4$: $\sin^{-1}\dfrac x4+C$.
4 Formula (6), $a=\sqrt7$: $\log\left|x+\sqrt{x^2+7}\right|+C$.
5 $\tfrac19\displaystyle\int\frac{dx}{x^2+\frac49}=\frac19\cdot\frac32\tan^{-1}\frac{3x}{2}+C$, that is, $\dfrac16\tan^{-1}\dfrac{3x}{2}+C$.
6 $\tfrac14\displaystyle\int\frac{dx}{\frac{25}4-x^2}$, formula (2) with $a=\tfrac52$: $\ \dfrac14\cdot\dfrac15\log\left|\dfrac{\frac52+x}{\frac52-x}\right|=\dfrac1{20}\log\left|\dfrac{5+2x}{5-2x}\right|+C$.
7 $2=A(1+x^2)+(Bx+C)(1-x)$: $x=1$ gives $A=1$; $x^2$: $0=A-B$, $B=1$; constant: $2=A+C$, $C=1$. So $\dfrac{1}{1-x}+\dfrac{x+1}{1+x^2}$ and the integral is $-\log|1-x|+\tfrac12\log(1+x^2)+\tan^{-1}x+C$.
8 $u=e^x$: $\displaystyle\int\frac{du}{(u-1)(u+2)}=\frac13\log\left|\frac{u-1}{u+2}\right|$, so the answer is $\tfrac13\log\left|\dfrac{e^x-1}{e^x+2}\right|+C$.
9 $u=\sin x$: $\dfrac{1}{(1-u)(2-u)}=\dfrac{1}{1-u}-\dfrac{1}{2-u}$, which integrates to $-\log|1-u|+\log|2-u|$. Answer: $\log\left|\dfrac{2-\sin x}{1-\sin x}\right|+C$.
Modelling. In each problem, time is the antiderivative of a rational function. Give answers to 2 decimal places, with units.
1Falling raindrop. A raindrop falls from rest and its acceleration is $9.8-0.2v^2$ m/s² at speed $v$ m/s, so the time to reach speed $v$ is $t(v)$ with $\dfrac{dt}{dv}=\dfrac{1}{9.8-0.2v^2}=\dfrac{5}{49-v^2}$ and $t(0)=0$. Find $t(v)$ and the time to reach $5$ m/s. What speed can the drop never exceed?
2Fish population. A lake is stocked with $10$ thousand fish, and the population $P$ (thousands) grows so that the time $t$ (years) to reach $P$ satisfies $\dfrac{dt}{dP}=\dfrac{200}{P(100-P)}$, with $t=0$ when $P=10$. Find $t(P)$ and the time for the population to reach $50$ thousand.
3Chemical reaction. In a reaction between two substances, the time $t$ (minutes) to form $x$ mol/L of product satisfies $\dfrac{dt}{dx}=\dfrac{10}{(3-x)(5-x)}$ for $0\le x\lt3$, with $t(0)=0$. Find $t(x)$ and the time to form $1$ mol/L.
💡 Show solutions▶
1 Formula (2) with $a=7$: $t=5\cdot\dfrac1{14}\log\left|\dfrac{7+v}{7-v}\right|+C$, and $t(0)=0$ gives $C=0$. So $t(v)=\dfrac{5}{14}\log\dfrac{7+v}{7-v}$ s and $t(5)=\dfrac5{14}\log6\approx0.64$ s. As $v\to7$, $t\to\infty$: the drop can never exceed its terminal speed $7$ m/s.
2 $\dfrac{200}{P(100-P)}=2\left(\dfrac1P+\dfrac{1}{100-P}\right)$, so $t=2\log\dfrac{P}{100-P}+C$. $t(10)=0$ gives $C=-2\log\tfrac19=2\log9$. Hence $t(P)=2\log\dfrac{9P}{100-P}$ years and $t(50)=2\log9\approx4.39$ years.
3 $\dfrac{1}{(3-x)(5-x)}=\dfrac12\left(\dfrac{1}{3-x}-\dfrac{1}{5-x}\right)$, so $t=5\big(-\log(3-x)+\log(5-x)\big)+C$, that is, $t=5\log\dfrac{5-x}{3-x}+C$. $t(0)=0$ gives $C=-5\log\tfrac53$. So $t(x)=5\log\dfrac{3(5-x)}{5(3-x)}$ min, and $t(1)=5\log\dfrac{12}{10}=5\log1.2\approx0.91$ min.
How this is examined: the Board almost always sets one partial-fractions integral (3 marks) and one “linear over quadratic” integral (often 5 marks), plus a 1- or 2-mark completing-the-square question. JEE likes substitutions that lead to partial fractions and questions on the constants in a decomposition.
A-part $\displaystyle\frac56\int\frac{6x+2}{3x^2+2x+1}\,dx=\frac56\log\left(3x^2+2x+1\right)$ (positive, since its discriminant is $4-12\lt0$). (1 mark)
Square $3x^2+2x+1=3\left[\left(x+\tfrac13\right)^2+\tfrac29\right]$, so with $a=\tfrac{\sqrt2}3$
$$\int\frac{dx}{3x^2+2x+1}=\frac13\cdot\frac{3}{\sqrt2}\tan^{-1}\frac{x+\frac13}{\frac{\sqrt2}3}=\frac{1}{\sqrt2}\tan^{-1}\frac{3x+1}{\sqrt2}$$ (2 marks)
Substitute Multiply above and below by $x^4$: $\displaystyle\int\frac{x^4\,dx}{x^5\left(x^5+1\right)}$. With $u=x^5$, $du=5x^4dx$: $\displaystyle\frac15\int\frac{du}{u(u+1)}$.
Answer (A). (B) and (D) mishandle the factor from $du=5x^4dx$; (C) has the fraction upside down (it is the negative of the answer).
⏱ Fast route For large $x$ the integrand is about $x^{-6}\gt0$, so the answer must increase; $\tfrac{x^5}{x^5+1}$ increases towards $1$, so only (A), (B), (D) are possible — and a factor $\tfrac15$ must come from $d(x^5)$.
E6.JEE Advanced(One or more options correct.) If $\displaystyle\int\frac{dx}{\left(x^2+1\right)\left(x^2+4\right)}$ $=A\tan^{-1}x+B\tan^{-1}\dfrac x2+C$, then (A) $A=\tfrac13$ (B) $B=-\tfrac16$ (C) $A+2B=0$ (D) $A-B=\tfrac16$
💡 Show solution▶
Split Treat $x^2$ as one unknown $t$: $\dfrac{1}{(t+1)(t+4)}=\dfrac13\left(\dfrac{1}{t+1}-\dfrac{1}{t+4}\right)$. So the integrand is $\dfrac13\cdot\dfrac1{x^2+1}-\dfrac13\cdot\dfrac{1}{x^2+4}$.
Integrate $\tfrac13\tan^{-1}x-\tfrac13\cdot\tfrac12\tan^{-1}\tfrac x2$, so $A=\tfrac13$ and $B=-\tfrac16$.
Answer (A), (B), (C). (D) is false: $A-B=\tfrac13+\tfrac16=\tfrac12$. The trap is to forget the $\tfrac1a=\tfrac12$ from $\int\frac{dx}{x^2+4}$, which would give $B=-\tfrac13$.
🔬 Research–Teaching NexusMini project · about 60 minutes
Research question. Does the spread of a new technology — mobile phones, internet connections, digital payments — follow the S-shaped logistic pattern, and when does it reach half its final level?
Why it matters. Planners at telecom regulators, banks and health ministries forecast adoption to decide how many towers, servers or vaccine doses are needed. If $P$ grows with rate $\dfrac{dP}{dt}=rP\left(1-\dfrac PK\right)$, then $\dfrac{dt}{dP}=\dfrac{K}{rP(K-P)}$, and partial fractions give
$$t=\frac1r\log\frac{P}{K-P}+C$$
So $\log\dfrac{P}{K-P}$ plotted against time should be a straight line of slope $r$ — a test you can do in a spreadsheet.
What to do.
Find yearly figures for an Indian adoption series (for example internet or broadband subscribers from TRAI reports, or digital-payment users, via data.gov.in). If you have no data, use the model values $P=4.0,\ 10.2,\ 23.5,\ 45.6,\ 69.5,\ 86.1$ (per cent of households) for years $t=0,2,4,6,8,10$.
Guess the saturation level $K$ (for percentages, try $K=100$). Compute $y=\log\dfrac{P}{K-P}$ for each year.
Plot $y$ against $t$ and fit a straight line. Its slope is $r$. Where does the line cross $y=0$? That is when $P=\tfrac K2$.
Try two other values of $K$. Which one makes the points most nearly straight?
Tools: Spreadsheet (LN, SLOPE, INTERCEPT, scatter chart with trendline).
Hand in: Your table, the straightened plot, your estimates of $r$ and of the half-way year, and a paragraph on whether the logistic model fits and what could make it fail.
Going further. Rearrange $t=\frac1r\log\frac{P}{K-P}+C$ to get $P$ as a function of $t$ and plot it over your data.
Instructor note: the model values come from $P=\dfrac{100}{1+24e^{-0.5t}}$ rounded to one decimal, so $y=\log\frac{P}{100-P}=-\log24+0.5t$: slope $r\approx0.5$ per year, intercept $\approx-3.18$, and $P=50$ at $t=2\log24\approx6.36$ years. A wrong $K$ bends the line — too small a $K$ makes it curve upwards at the end.
🧪 Self-check quiz
0/5 correct
Q1. $\displaystyle\int\frac{dx}{x^2+9}$ equals
Q2. Completing the square, $x^2-6x+13$ equals
Q3. The correct form of the partial fractions of $\dfrac{2x+3}{(x-1)^2(x+2)}$ is