Quick Recall
Quick check — pick the correct statement before we meet today's wall.
$\displaystyle\int \frac{1}{ax+b}\, \text{d}x = \frac{1}{a}\ln|ax+b| + C$, because $\dfrac{\text{d}}{\text{d}x}\big(\ln|ax+b|\big) = \dfrac{a}{ax+b}$ and the extra factor $a$ must be cancelled.
$\dfrac{\text{d}}{\text{d}x}\big(\ln|ax+b|\big) = \dfrac{a}{ax+b}$ — the inner function $ax+b$ differentiates to $a$. Reversing this to integrate $\dfrac{1}{ax+b}$ therefore needs a compensating $\dfrac{1}{a}$. Today's lesson generalises this same "reverse the chain rule" idea.
The Wall
But now consider this integral:
$$\int \frac{x}{x^2+1}\, \text{d}x = \;?$$
The denominator is $x^2+1$, not a simple $ax+b$. The formula from L1 does not directly apply. A student tries “maybe $\ln|x^2+1| + C$?” — but differentiating $\ln|x^2+1|$ gives $\dfrac{2x}{x^2+1}$ (chain rule!), which is twice the integrand. The guess is too big by a factor of 2.
The Need
The key insight: the derivative of the denominator is $2x$, and the numerator $x$ is exactly half of $2x$. So we can write:
$$\frac{x}{x^2+1} = \frac{1}{2} \cdot \frac{2x}{x^2+1} = \frac{1}{2} \cdot \frac{f'(x)}{f(x)}$$
where $f(x) = x^2+1$ and $f'(x) = 2x$.
Today we learn the general rule: if the numerator is a constant multiple of the derivative of the denominator, the integral is a logarithm.
💡 The Bridge (click to reveal)
The chain rule for logarithms says: $\dfrac{\text{d}}{\text{d}x}\!\left[\ln|f(x)|\right] = \dfrac{f'(x)}{f(x)}$.
Reversing this gives the master formula:
$$\int \frac{f'(x)}{f(x)}\, \text{d}x = \ln|f(x)| + C$$
More generally, if the numerator is $k$ times the derivative of the denominator:
$$\int \frac{k\,f'(x)}{f(x)}\, \text{d}x = k\ln|f(x)| + C$$
For our problem: $\displaystyle\int \frac{x}{x^2+1}\, \text{d}x = \frac{1}{2}\ln|x^2+1| + C$. Verify: $\dfrac{\text{d}}{\text{d}x}\!\left(\frac{1}{2}\ln|x^2+1|\right) = \frac{1}{2} \cdot \frac{2x}{x^2+1} = \frac{x}{x^2+1}$. It checks!
Last lesson we used partial fractions to integrate rational functions. Today we learn to recognise a special pattern: when the numerator is a constant multiple of the derivative of the denominator, the integral is a logarithm.
From the chain rule, $\dfrac{\text{d}}{\text{d}x}\!\left[\ln|f(x)|\right] = \dfrac{f'(x)}{f(x)}$. Reversing gives:
$$\boxed{\;\int \frac{k\,f'(x)}{f(x)}\, \text{d}x = k\ln|f(x)| + C\;}\qquad(f(x) \neq 0)$$The key test: is the numerator a constant multiple of the derivative of the denominator? If yes, identify $f(x)$, $f'(x)$, and the constant $k$, then apply the formula.
This is a special case of the master pattern with $f(x) = ax+b$, $f'(x) = a$. Here $\dfrac{1}{ax+b} = \dfrac{1}{a} \cdot \dfrac{a}{ax+b} = \dfrac{1}{a} \cdot \dfrac{f'(x)}{f(x)}$, so $k = \dfrac{1}{a}$:
$$\boxed{\;\int \frac{1}{ax+b}\, \text{d}x = \frac{1}{a}\ln|ax+b| + C\;}\qquad(a \neq 0)$$This connects back to Lesson 1 — the formula we already knew is just the $kf'(x)/f(x)$ pattern with $f(x) = ax+b$.
Since $\tan x = \dfrac{\sin x}{\cos x}$, let $f(x) = \cos x$. Then $f'(x) = -\sin x$, so $\sin x = -f'(x)$.
Therefore $\tan x = \dfrac{\sin x}{\cos x} = \dfrac{-f'(x)}{f(x)} = -\dfrac{f'(x)}{f(x)}$, meaning $k = -1$.
Applying the master pattern:
$$\boxed{\;\int \tan x\, \text{d}x = -\ln|\cos x| + C\;}\qquad(\cos x \neq 0)$$This is one of the most important results in the integration toolkit. Verify: $\dfrac{\text{d}}{\text{d}x}\!\left(-\ln|\cos x|\right) = -\dfrac{-\sin x}{\cos x} = \tan x$. It checks!
In groups of three you must approve each line before it is shown. At every stage, pick the line you believe comes next in the evaluation of $\displaystyle\int \dfrac{4x+6}{x^{2}+3x+5}\,\text{d}x$. A correct pick reveals the line and unlocks the next stage; a wrong pick turns red and explains the trap. Four or fewer wrong picks in the whole run earns the top band.
Group rule: nobody clicks until all three members agree on the line. If you disagree, say why — that disagreement is where the learning happens.
On your own. Each level shows three integrals. Your job is to name the rule they share, not to evaluate them. Level 2 only opens when level 1 is cleared, and only a perfect first attempt at every level earns all three stars.
Find $\displaystyle\int \frac{2x+3}{3x^{2}+9x-1} \, \text{d}x$.
👁 Reveal Solution
The denominator plays the part of $f(x)$, so take $f(x) = 3x^{2}+9x-1$.
Differentiate it: $f'(x) = 6x+9 = 3(2x+3)$.
Now test the numerator. Is $2x+3$ a constant multiple of $f'(x)$? Yes:
$2x+3 = \dfrac{1}{3}\cdot 3(2x+3) = \dfrac{1}{3}f'(x)$, so $k = \dfrac{1}{3}$, which is a constant.
Since the numerator is a constant multiple of the derivative of the denominator, the master pattern applies. (If the multiplier depended on $x$ it would not be a constant, and the pattern would fail.)
Using $\displaystyle\int \frac{k\,f'(x)}{f(x)}\, \text{d}x = k\ln|f(x)| + C$ with $k = \dfrac{1}{3}$:
$\displaystyle\int \frac{2x+3}{3x^{2}+9x-1}\, \text{d}x = \frac{1}{3}\ln|3x^{2}+9x-1| + C$
$\boxed{\displaystyle\int \frac{2x+3}{3x^{2}+9x-1}\, \text{d}x = \frac{1}{3}\ln|3x^{2}+9x-1| + C}$
$\dfrac{\text{d}}{\text{d}x}\!\left[\frac{1}{3}\ln|3x^{2}+9x-1|\right] = \frac{1}{3}\cdot\dfrac{6x+9}{3x^{2}+9x-1} = \dfrac{2x+3}{3x^{2}+9x-1}$. It checks!
The integral is valid wherever $3x^{2}+9x-1 \neq 0$.
Find $\displaystyle\int_{4}^{6} \frac{x-3}{x^{2}-6x+9} \, \text{d}x$, giving your answer to 3 significant figures where appropriate.
👁 Reveal Solution
Write the denominator as a perfect square: $x^{2}-6x+9 = (x-3)^{2}$, so take $f(x) = (x-3)^{2}$.
Then $f'(x) = 2(x-3) = 2x-6$, and the numerator is $x-3 = \dfrac{1}{2}f'(x)$. So $k = \dfrac{1}{2}$.
$\displaystyle\int \frac{x-3}{x^{2}-6x+9}\, \text{d}x = \frac{1}{2}\ln\left|(x-3)^{2}\right| + C$
Using $\ln|a^{2}| = 2\ln|a|$, this is $\dfrac{1}{2}\cdot 2\ln|x-3| = \ln|x-3| + C$.
$\displaystyle\int_{4}^{6} \frac{x-3}{x^{2}-6x+9}\, \text{d}x = \left[\frac{1}{2}\ln\left|(x-3)^{2}\right|\right]_{4}^{6}$
$= \frac{1}{2}\left(\ln\left|(6-3)^{2}\right| - \ln\left|(4-3)^{2}\right|\right) = \frac{1}{2}\left(\ln 9 - \ln 1\right)$
$= \frac{1}{2}\ln 9 = \frac{1}{2} \cdot 2\ln 3 = \ln 3 \approx 1.10$ (3 s.f.)
$\boxed{\displaystyle\int_{4}^{6} \frac{x-3}{x^{2}-6x+9}\, \text{d}x = \ln 3 \approx 1.10}$
On $4 \leqslant x \leqslant 6$ we have $x-3 > 0$, so the modulus signs are not needed in the final answer.
The denominator is zero only at $x = 3$, which lies outside the interval of integration, so the definite integral is well defined.
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Find the value of $\displaystyle\int_{1}^{\mathrm{e}} \frac{2t}{t^{2}+1} \, \text{d}t$.
💡 Hint
Take $f(t) = t^{2}+1$, so $f'(t) = 2t$. The numerator is exactly $f'(t)$, so $k = 1$.
The antiderivative is $\ln|t^{2}+1|$. Evaluate at $t = 1$ and $t = \mathrm{e}$, then simplify with log laws.
👁 Reveal Solution
With $f(t) = t^{2}+1$ we have $f'(t) = 2t$, so $\dfrac{2t}{t^{2}+1} = \dfrac{f'(t)}{f(t)}$ and $k = 1$.
$\displaystyle\int \frac{2t}{t^{2}+1}\, \text{d}t = \ln|t^{2}+1| + C$
$\displaystyle\int_{1}^{\mathrm{e}} \frac{2t}{t^{2}+1}\, \text{d}t = \left[\ln(t^{2}+1)\right]_{1}^{\mathrm{e}}$
$= \ln(\mathrm{e}^{2}+1) - \ln(1^{2}+1) = \ln(\mathrm{e}^{2}+1) - \ln 2$
$= \ln\!\left(\dfrac{\mathrm{e}^{2}+1}{2}\right) \approx 1.43$
$\boxed{\displaystyle\int_{1}^{\mathrm{e}} \frac{2t}{t^{2}+1}\, \text{d}t = \ln\!\left(\frac{\mathrm{e}^{2}+1}{2}\right) \approx 1.43}$
The curve $y = \dfrac{\mathrm{e}^{x}}{\mathrm{e}^{x}+1}$ lies above the $x$-axis for all $x$.
Find the area of the region enclosed by this curve, the axes and the line $x = 2$.
💡 Hint
The region is bounded on the left by the $y$-axis ($x = 0$) and on the right by $x = 2$, so the area is $\displaystyle\int_{0}^{2} \frac{\mathrm{e}^{x}}{\mathrm{e}^{x}+1}\, \text{d}x$.
Take $f(x) = \mathrm{e}^{x}+1$. Then $f'(x) = \mathrm{e}^{x}$, which is exactly the numerator, so $k = 1$ and the antiderivative is $\ln(\mathrm{e}^{x}+1)$.
👁 Reveal Solution
The region is bounded by the curve, the $x$-axis, the $y$-axis and the line $x = 2$, so
$\text{Area} = \displaystyle\int_{0}^{2} \frac{\mathrm{e}^{x}}{\mathrm{e}^{x}+1}\, \text{d}x$
Take $f(x) = \mathrm{e}^{x}+1$, so $f'(x) = \mathrm{e}^{x}$. The numerator is $f'(x)$ itself, so $k = 1$:
$\displaystyle\int \frac{\mathrm{e}^{x}}{\mathrm{e}^{x}+1}\, \text{d}x = \ln|\mathrm{e}^{x}+1| + C = \ln(\mathrm{e}^{x}+1) + C$
Here $\mathrm{e}^{x}+1 > 0$ for all $x$, so the modulus signs may be dropped.
$\displaystyle\int_{0}^{2} \frac{\mathrm{e}^{x}}{\mathrm{e}^{x}+1}\, \text{d}x = \left[\ln(\mathrm{e}^{x}+1)\right]_{0}^{2}$
$= \ln(\mathrm{e}^{2}+1) - \ln(\mathrm{e}^{0}+1) = \ln(\mathrm{e}^{2}+1) - \ln 2$
$= \ln\!\left(\dfrac{\mathrm{e}^{2}+1}{2}\right) \approx 1.43$
$\boxed{\text{Area} = \ln\!\left(\frac{\mathrm{e}^{2}+1}{2}\right) \approx 1.43}$
(i) The curve $y = \dfrac{x}{x^{2}+1}$ has a maximum point $M$. Find the $x$ coordinate of $M$.
(ii) The region $R$ is bounded by the curve, the $x$-axis and the line $x = p$. Find the area of $R$ in terms of $p$.
(iii) Hence calculate the value of $p$ for which the area of $R$ is 1, giving your answer correct to 3 significant figures.
(9709 Paper 3 Q9 June 2005)
💡 Hint 1
For (i): differentiate $y = \dfrac{x}{x^{2}+1}$ with the quotient rule, then solve $\dfrac{\text{d}y}{\text{d}x} = 0$.
Two values of $x$ satisfy that equation. Decide which one gives the maximum by checking the sign of $\dfrac{\text{d}y}{\text{d}x}$ on either side, or by comparing the two $y$ values.
💡 Hint 2
For (ii): the region runs from $x = 0$ to $x = p$, so $\text{Area} = \displaystyle\int_{0}^{p} \frac{x}{x^{2}+1}\, \text{d}x$. This is the $kf'(x)/f(x)$ pattern with $f(x) = x^{2}+1$ and $f'(x) = 2x$, so $k = \dfrac{1}{2}$.
For (iii): set your expression for the area equal to 1, then use $\mathrm{e}^{\ln a} = a$ to remove the logarithm.
👁 Reveal Solution
By the quotient rule, with $u = x$ and $v = x^{2}+1$:
$\dfrac{\text{d}y}{\text{d}x} = \dfrac{(1)(x^{2}+1) - x(2x)}{(x^{2}+1)^{2}} = \dfrac{1-x^{2}}{(x^{2}+1)^{2}}$
Setting $\dfrac{\text{d}y}{\text{d}x} = 0$ gives $1-x^{2} = 0$, so $x = 1$ or $x = -1$.
The denominator $(x^{2}+1)^{2}$ is always positive, so the sign of $\dfrac{\text{d}y}{\text{d}x}$ is the sign of $1-x^{2}$: positive for $-1 < x < 1$ and negative for $x > 1$.
The curve therefore rises up to $x = 1$ and falls afterwards, so the maximum point $M$ has $x = 1$. (At $x = -1$ we get $y = -\dfrac{1}{2}$, a minimum.)
$\boxed{x = 1}$
The curve is above the $x$-axis for $x > 0$, so $R$ runs from $x = 0$ to $x = p$ and
$\text{Area} = \displaystyle\int_{0}^{p} \frac{x}{x^{2}+1}\, \text{d}x$
Here $f(x) = x^{2}+1$, $f'(x) = 2x$ and $x = \dfrac{1}{2}f'(x)$, so $k = \dfrac{1}{2}$. Hence
$\text{Area} = \left[\frac{1}{2}\ln(x^{2}+1)\right]_{0}^{p} = \frac{1}{2}\ln(p^{2}+1) - \frac{1}{2}\ln 1 = \frac{1}{2}\ln(1+p^{2})$
$\boxed{\text{Area of }R = \frac{1}{2}\ln(1+p^{2})}$
Set $\dfrac{1}{2}\ln(1+p^{2}) = 1$:
$\ln(1+p^{2}) = 2 \quad\Longrightarrow\quad 1+p^{2} = \mathrm{e}^{2}$
$p^{2} = \mathrm{e}^{2}-1 \quad\Longrightarrow\quad p = \sqrt{\mathrm{e}^{2}-1} = 2.5277\ldots$
Since $p > 0$ we take the positive root, giving $p = 2.53$ (3 s.f.).
$\boxed{p = \sqrt{\mathrm{e}^{2}-1} = 2.53 \text{ (3 s.f.)}}$
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.