Quick Recall
Quick check — pick the correct statement in each question before we meet today's wall.
$\displaystyle\int u v' \, \text{d}x = uv - \int u' v \, \text{d}x$ — integrate the product rule $(uv)' = u'v + uv'$ and rearrange. The minus sign before the second integral is easy to drop; watch for it.
LIATE order is Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential. $\ln x$ is Logarithmic and $x^2$ is Algebraic, so choose $u=\ln x$; differentiating it removes the $\ln$ while integrating $x^2$ is easy.
The Wall
Now consider this integral:
$$\int \ln x \, \text{d}x$$
There is only one function here: $\ln x$. The by parts formula needs two factors — one to differentiate ($u$) and one to integrate ($v'$). How can you apply a two-factor formula to a one-factor integral?
Similarly, what about $\int \tan^{-1} x \, \text{d}x$? Again, only one factor — and we cannot integrate $\tan^{-1} x$ directly.
The Need
We need a systematic technique for integrals that appear to have only one factor — logarithms, inverse trig functions, and composite functions like $(\ln x)^2$ or $\sin(\ln x)$. The trick: treat the integrand as $f(x) \cdot 1$, choose $u = f(x)$ and $v' = 1$, then apply by parts. Today we also meet cyclic integration by parts, where the original integral reappears and we solve for it algebraically.
💡 The Bridge (click to reveal)
Write $\int \ln x \, \text{d}x = \int \underbrace{\ln x}_{u} \cdot \underbrace{1}_{v'} \, \text{d}x$. By LIATE, $\ln x$ (Logarithmic, first priority) is $u$, and $1$ integrates to $v = x$.
Then $u' = \dfrac{1}{x}$ and $v = x$, so:
$$\int \ln x \, \text{d}x = x \ln x - \int x \cdot \frac{1}{x} \, \text{d}x = x \ln x - x + C$$
The same trick works for $\int \tan^{-1} x \, \text{d}x$ — set $u = \tan^{-1} x$, $v' = 1$. And for cyclic integrals like $\int e^x \sin x \, \text{d}x$, applying by parts twice brings the original integral back — we then solve for it like an equation.
In Lesson 1, we applied integration by parts to products of two visible factors. Today we extend it to special cases: integrals with only one factor (where we reveal a hidden “$1$”), and cyclic integrals where the original integral reappears after two applications.
When the integrand is a single function $f(x)$ that cannot be integrated directly but whose derivative $f'(x)$ is simpler, rewrite:
$$\int f(x) \, \text{d}x = \int f(x) \cdot 1 \, \text{d}x = \int \underbrace{f(x)}_{u} \cdot \underbrace{1}_{v'} \, \text{d}x$$
Then $u' = f'(x)$ and $v = x$. Applying by parts: $\int f(x) \, \text{d}x = x \cdot f(x) - \int x \cdot f'(x) \, \text{d}x$. The new integral $\int x \cdot f'(x) \, \text{d}x$ is often simpler because $f'(x)$ has lower complexity than $f(x)$.
For integrals like $\int e^x \sin x \, \text{d}x$ or $\int e^x \cos x \, \text{d}x$, applying by parts twice brings back the original integral. Let $I = \int e^x \sin x \, \text{d}x$. After two applications:
$$I = e^x \sin x - e^x \cos x - I$$
The original $I$ reappears on the right! Solve algebraically: $2I = e^x(\sin x - \cos x)$, so $I = \dfrac{1}{2} e^x (\sin x - \cos x) + C$. The key sign: keep the same choice of $u$ in both applications (e.g. always $u = $ trig, $v' = e^x$).
The LIATE hierarchy (Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential) tells us which factor to choose as $u$. For special cases with a hidden $1$:
- $\int \ln x \, \text{d}x$: $u = \ln x$ (L), $v' = 1$ — L is first in LIATE.
- $\int \tan^{-1} x \, \text{d}x$: $u = \tan^{-1} x$ (I), $v' = 1$ — I is second in LIATE.
- $\int (\ln x)^2 \, \text{d}x$: $u = (\ln x)^2$ (L), $v' = 1$ — still L first.
The pattern: when one factor is $1$ (Algebraic), the other factor is always higher in LIATE and becomes $u$.
- If the integrand is a single function, rewrite it as $f(x) \cdot 1$ and use $u = f(x)$, $v' = 1$.
- Apply by parts: $\int f(x) \, \text{d}x = x \cdot f(x) - \int x \cdot f'(x) \, \text{d}x$.
- If the new integral is still a product, apply by parts again.
- If the original integral reappears, call it $I$ and solve algebraically: $2I = \ldots$
- Always verify by differentiating your answer.
Work in a group of four. The board solves $\displaystyle\int \ln x\,\text{d}x$ by revealing the hidden factor, and each stage belongs to one role — Chooser, Differentiator, Assembler, Finisher. Only the owner of a stage may click, and no stage unlocks until the stage above it is right, so the group must agree on every line before it moves. Each option is a judgement: say out loud why the other two fail.
Role rule: the owner explains the reasoning out loud before anyone clicks — the rest of the group may only ask "why?".
Two worked by-parts solutions, 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 — no guessing.
Find $\displaystyle\int \ln x \, \text{d}x$ using integration by parts.
(This integral has only one factor. Use the $v' = 1$ trick: rewrite $\ln x = \ln x \cdot 1$.)
👁 Reveal Solution
Write $\displaystyle\int \ln x \, \text{d}x = \int \underbrace{\ln x}_{u} \cdot \underbrace{1}_{v'} \, \text{d}x$.
By LIATE: $u = \ln x$ (Logarithmic), $v' = 1$ (integrate the constant $1$).
$u' = \dfrac{1}{x}$, $v = \displaystyle\int \text{d}x = x$.
$\displaystyle\int \ln x \, \text{d}x = uv - \int u' v \, \text{d}x = (\ln x)(x) - \int x \cdot \frac{1}{x} \, \text{d}x = x \ln x - \int 1 \, \text{d}x$.
$\boxed{\displaystyle\int \ln x \, \text{d}x = x \ln x - x + C}$
Verification: $\dfrac{\text{d}}{\text{d}x}(x \ln x - x) = \ln x + x \cdot \frac{1}{x} - 1 = \ln x + 1 - 1 = \ln x$. ✓
Find $\displaystyle\int \tan^{-1} x \, \text{d}x$ using integration by parts.
(Use the same $v' = 1$ trick: $u = \tan^{-1} x$, $v' = 1$. Recall that $\dfrac{\text{d}}{\text{d}x} \tan^{-1} x = \dfrac{1}{1+x^2}$.)
👁 Reveal Solution
$\displaystyle\int \tan^{-1} x \, \text{d}x = \int \underbrace{\tan^{-1} x}_{u} \cdot \underbrace{1}_{v'} \, \text{d}x$.
$u = \tan^{-1} x$ (Inverse trig), $v' = 1$.
$u' = \dfrac{1}{1+x^2}$, $v = x$.
$\displaystyle\int \tan^{-1} x \, \text{d}x = x \tan^{-1} x - \int \frac{x}{1+x^2} \, \text{d}x$.
For the remaining integral, recognise the log form: $\dfrac{\text{d}}{\text{d}x}\ln(1+x^2) = \dfrac{2x}{1+x^2}$, so reversing the derivative gives $\displaystyle\int \frac{2x}{1+x^2} \, \text{d}x = \ln(1+x^2)$.
Hence $\displaystyle\int \frac{x}{1+x^2} \, \text{d}x = \frac{1}{2} \int \frac{2x}{1+x^2} \, \text{d}x = \frac{1}{2} \ln(1+x^2)$.
$\boxed{\displaystyle\int \tan^{-1} x \, \text{d}x = x \tan^{-1} x - \tfrac{1}{2}\ln(1+x^2) + C}$
Verification: $\dfrac{\text{d}}{\text{d}x}\!\left[x \tan^{-1} x - \tfrac{1}{2}\ln(1+x^2)\right] = \tan^{-1} x + \dfrac{x}{1+x^2} - \dfrac{x}{1+x^2} = \tan^{-1} x$. ✓
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Find $\displaystyle\int (\ln x)^2 \, \text{d}x$.
(Use the $v' = 1$ trick with $u = (\ln x)^2$, $v' = 1$. The result will involve $\int \ln x \, \text{d}x$, which you already know.)
💡 Hint
Let $u = (\ln x)^2$, $v' = 1$. Then $u' = 2(\ln x) \cdot \frac{1}{x}$ (chain rule) and $v = x$.
Apply by parts: $\int (\ln x)^2 \, \text{d}x = x(\ln x)^2 - \int x \cdot 2 \frac{\ln x}{x} \text{d}x = x(\ln x)^2 - 2\int \ln x \, \text{d}x$.
Use the known result $\int \ln x \, \text{d}x = x \ln x - x + C$.
👁 Reveal Solution
$u = (\ln x)^2$ (Logarithmic), $v' = 1$.
$u' = 2(\ln x) \cdot \dfrac{1}{x}$ (chain rule), $v = x$.
$\displaystyle\int (\ln x)^2 \, \text{d}x = x(\ln x)^2 - \int x \cdot 2\frac{\ln x}{x} \, \text{d}x = x(\ln x)^2 - 2\int \ln x \, \text{d}x$.
$\displaystyle\int (\ln x)^2 \, \text{d}x = x(\ln x)^2 - 2(x \ln x - x) + C$.
$\boxed{x(\ln x)^2 - 2x \ln x + 2x + C}$
Verification: $\dfrac{\text{d}}{\text{d}x}\!\left[x(\ln x)^2 - 2x \ln x + 2x\right] = (\ln x)^2 + 2\ln x - 2\ln x - 2 + 2 = (\ln x)^2$. ✓
Find the exact value of $\displaystyle\int_1^2 x^2 \ln x \, \text{d}x$.
(In L1 you found the indefinite integral $\int x^2 \ln x \, \text{d}x$. Here we extend to a definite integral.)
💡 Hint
Use LIATE: $u = \ln x$ (Logarithmic), $v' = x^2$ (Algebraic).
Then $u' = \frac{1}{x}$ and $v = \frac{x^3}{3}$. Apply by parts, then evaluate at $x = 2$ and $x = 1$.
👁 Reveal Solution
$u = \ln x$ (Logarithmic), $v' = x^2$ (Algebraic).
$u' = \dfrac{1}{x}$, $v = \dfrac{x^3}{3}$.
$\displaystyle\int x^2 \ln x \, \text{d}x = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, \text{d}x = \frac{x^3}{3} \ln x - \frac{1}{3}\int x^2 \, \text{d}x = \frac{x^3}{3} \ln x - \frac{x^3}{9} + C$.
$\displaystyle\int_1^2 x^2 \ln x \, \text{d}x = \left[\frac{x^3}{3} \ln x - \frac{x^3}{9}\right]_1^2 = \left(\frac{8}{3}\ln 2 - \frac{8}{9}\right) - \left(\frac{1}{3}\ln 1 - \frac{1}{9}\right)$.
Since $\ln 1 = 0$: $= \dfrac{8}{3}\ln 2 - \dfrac{8}{9} + \dfrac{1}{9} = \dfrac{8}{3}\ln 2 - \dfrac{7}{9}$.
$\boxed{\displaystyle\int_1^2 x^2 \ln x \, \text{d}x = \frac{8}{3}\ln 2 - \frac{7}{9}}$
Numerical check: $\frac{8}{3}(0.6931) - 0.7778 \approx 1.8484 - 0.7778 = 1.0706$. By numerical integration: $\int_1^2 x^2 \ln x \, \text{d}x \approx 1.0706$. ✓
Find $\displaystyle\int e^x \sin x \, \text{d}x$.
(This is a cyclic integration by parts problem. Apply by parts twice — the original integral will reappear. Keep the same type for $u$ in both applications.)
💡 Hint 1
Let $I = \int e^x \sin x \, \text{d}x$. Choose $u = \sin x$ (Trigonometric), $v' = e^x$ (Exponential).
Then $u' = \cos x$ and $v = e^x$. Apply by parts once.
💡 Hint 2
After the first application: $I = e^x \sin x - \int e^x \cos x \, \text{d}x$.
Apply by parts again to $\int e^x \cos x \, \text{d}x$ with $u = \cos x$ (same type: trig), $v' = e^x$. The resulting integral will contain $I$ again!
👁 Reveal Solution
Let $I = \displaystyle\int e^x \sin x \, \text{d}x$. Choose $u = \sin x$, $v' = e^x$.
$u' = \cos x$, $v = e^x$.
$I = e^x \sin x - \displaystyle\int e^x \cos x \, \text{d}x$.
Choose $u = \cos x$ (same type: trig), $v' = e^x$.
$u' = -\sin x$, $v = e^x$.
$\displaystyle\int e^x \cos x \, \text{d}x = e^x \cos x - \int e^x(-\sin x) \, \text{d}x = e^x \cos x + \int e^x \sin x \, \text{d}x = e^x \cos x + I$.
$I = e^x \sin x - (e^x \cos x + I) = e^x \sin x - e^x \cos x - I$.
$2I = e^x \sin x - e^x \cos x = e^x(\sin x - \cos x)$.
$I = \dfrac{1}{2} e^x (\sin x - \cos x) + C$.
$\boxed{\displaystyle\int e^x \sin x \, \text{d}x = \tfrac{1}{2} e^x(\sin x - \cos x) + C}$
Verification: $\dfrac{\text{d}}{\text{d}x}\!\left[\tfrac{1}{2} e^x(\sin x - \cos x)\right] = \tfrac{1}{2}\left[e^x(\sin x - \cos x) + e^x(\cos x + \sin x)\right] = \tfrac{1}{2} e^x(2\sin x) = e^x \sin x$. ✓
Find $\displaystyle\int \sin(\ln x) \, \text{d}x$.
(This combines the $v' = 1$ trick with cyclic by parts. Let $I = \int \sin(\ln x) \, \text{d}x$, set $u = \sin(\ln x)$, $v' = 1$, and apply by parts twice.)
💡 Hint 1
Let $I = \int \sin(\ln x) \, \text{d}x$. Use the $v' = 1$ trick: $u = \sin(\ln x)$, $v' = 1$.
Then $u' = \cos(\ln x) \cdot \frac{1}{x}$ (chain rule) and $v = x$. Apply by parts once.
💡 Hint 2
After the first application: $I = x \sin(\ln x) - \int x \cdot \cos(\ln x) \cdot \frac{1}{x} \text{d}x = x \sin(\ln x) - \int \cos(\ln x) \, \text{d}x$.
Let $J = \int \cos(\ln x) \, \text{d}x$. Apply by parts again with $u = \cos(\ln x)$, $v' = 1$. The result will contain $I$ again!
👁 Reveal Solution
Let $I = \displaystyle\int \sin(\ln x) \, \text{d}x$. Choose $u = \sin(\ln x)$, $v' = 1$.
$u' = \cos(\ln x) \cdot \dfrac{1}{x}$ (chain rule), $v = x$.
$I = x \sin(\ln x) - \displaystyle\int x \cdot \cos(\ln x) \cdot \frac{1}{x} \, \text{d}x = x \sin(\ln x) - \int \cos(\ln x) \, \text{d}x$.
Choose $u = \cos(\ln x)$, $v' = 1$.
$u' = -\sin(\ln x) \cdot \dfrac{1}{x}$, $v = x$.
$J = x \cos(\ln x) - \displaystyle\int x \cdot \left(-\sin(\ln x) \cdot \frac{1}{x}\right) \text{d}x = x \cos(\ln x) + \int \sin(\ln x) \, \text{d}x = x \cos(\ln x) + I$.
$I = x \sin(\ln x) - J = x \sin(\ln x) - (x \cos(\ln x) + I) = x \sin(\ln x) - x \cos(\ln x) - I$.
$2I = x \sin(\ln x) - x \cos(\ln x) = x\bigl(\sin(\ln x) - \cos(\ln x)\bigr)$.
$I = \dfrac{x}{2}\bigl(\sin(\ln x) - \cos(\ln x)\bigr) + C$.
$\boxed{\displaystyle\int \sin(\ln x) \, \text{d}x = \tfrac{x}{2}\bigl(\sin(\ln x) - \cos(\ln x)\bigr) + C}$
Verification: $\dfrac{\text{d}}{\text{d}x}\!\left[\tfrac{x}{2}(\sin(\ln x) - \cos(\ln x))\right] = \tfrac{1}{2}(\sin(\ln x) - \cos(\ln x)) + \tfrac{x}{2}\!\left(\cos(\ln x) \cdot \frac{1}{x} + \sin(\ln x) \cdot \frac{1}{x}\right) = \tfrac{1}{2}\sin(\ln x) - \tfrac{1}{2}\cos(\ln x) + \tfrac{1}{2}\cos(\ln x) + \tfrac{1}{2}\sin(\ln x) = \sin(\ln x)$. ✓
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.