| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Parts |
| Type | Integration of x^n·ln(x) |
| Difficulty | Standard +0.3 This is a straightforward integration by parts question with clear guidance. Part (i) provides a helpful hint by asking for the derivative first, and part (ii) explicitly tells students to use integration by parts. The integration itself is standard (choosing u = ln x, dv = x^{-2}dx) with no conceptual surprises, making it slightly easier than the average A-level question which typically requires more independent decision-making. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Quotient rule with \(u = \ln x\) and \(v = x^2\) | M1 | \(udv \pm vdu\) in quotient rule is M0 |
| \(\frac{d}{dx}(\ln x) = 1/x\) soi | B1 | |
| \(\frac{dy}{dx} = \frac{x^2 \cdot \frac{1}{x} - \ln x \cdot 2x}{x^4}\) correct expression | A1 | Condone \(\ln x \cdot 2x = \ln 2x^2\) for this A1 (provided \(\ln x \cdot 2x\) is shown) |
| \(= \frac{1 - 2\ln x}{x^3}\) | A1 [4] | o.e. cao; mark final answer; must have divided top and bottom by \(x\); e.g. \(\frac{1}{x^3} - \frac{2\ln x}{x^3}\), \(x^{-3} - 2x^{-3}\ln x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = -2x^{-3}\ln x + x^{-2}\cdot\frac{1}{x}\) | M1, B1, A1 | Product rule; \(d/dx(\ln x)=1/x\) soi; correct expression |
| \(= -2x^{-3}\ln x + x^{-3}\) | A1 [4] | cao; must simplify the \(x^{-2}\cdot(1/x)\) term |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Integration by parts with \(u = \ln x\), \(du/dx = 1/x\), \(dv/dx = 1/x^2\), \(v = -x^{-1}\) | M1 | Must be correct |
| \(= -\frac{1}{x}\ln x + \int \frac{1}{x}\cdot\frac{1}{x}\,dx\) | A1 | Must be correct; condone \(+c\); need to see \(1/x^2\) at this stage |
| \(= -\frac{1}{x}\ln x - \frac{1}{x} + c\) | A1 | Condone missing \(c\) |
| \(= -\frac{1}{x}(\ln x + 1) + c\) | A1 [4] | NB AG; must have \(c\) shown in final answer |
## Question 3:
### Part (i): $\frac{d}{dx}\left(\frac{\ln x}{x^2}\right)$
| Answer/Working | Marks | Guidance |
|---|---|---|
| Quotient rule with $u = \ln x$ and $v = x^2$ | M1 | $udv \pm vdu$ in quotient rule is M0 |
| $\frac{d}{dx}(\ln x) = 1/x$ soi | B1 | |
| $\frac{dy}{dx} = \frac{x^2 \cdot \frac{1}{x} - \ln x \cdot 2x}{x^4}$ correct expression | A1 | Condone $\ln x \cdot 2x = \ln 2x^2$ for this A1 (provided $\ln x \cdot 2x$ is shown) |
| $= \frac{1 - 2\ln x}{x^3}$ | A1 [4] | o.e. cao; mark final answer; must have divided top and bottom by $x$; e.g. $\frac{1}{x^3} - \frac{2\ln x}{x^3}$, $x^{-3} - 2x^{-3}\ln x$ |
**Or** using product rule with $u = x^{-2}$ and $v = \ln x$:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -2x^{-3}\ln x + x^{-2}\cdot\frac{1}{x}$ | M1, B1, A1 | Product rule; $d/dx(\ln x)=1/x$ soi; correct expression |
| $= -2x^{-3}\ln x + x^{-3}$ | A1 [4] | cao; must simplify the $x^{-2}\cdot(1/x)$ term |
### Part (ii): $\int \frac{\ln x}{x^2}\,dx$
| Answer/Working | Marks | Guidance |
|---|---|---|
| Integration by parts with $u = \ln x$, $du/dx = 1/x$, $dv/dx = 1/x^2$, $v = -x^{-1}$ | M1 | Must be correct |
| $= -\frac{1}{x}\ln x + \int \frac{1}{x}\cdot\frac{1}{x}\,dx$ | A1 | Must be correct; condone $+c$; need to see $1/x^2$ at this stage |
| $= -\frac{1}{x}\ln x - \frac{1}{x} + c$ | A1 | Condone missing $c$ |
| $= -\frac{1}{x}(\ln x + 1) + c$ | A1 [4] | **NB AG**; must have $c$ shown in final answer |
---3 (i) Differentiate $\frac { \ln x } { x ^ { 2 } }$, simplifying your answer.\\
(ii) Using integration by parts, show that $\int \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x = - \frac { 1 } { x } ( 1 + \ln x ) + c$.
\hfill \mbox{\textit{OCR MEI C3 Q3 [8]}}