| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Indefinite integral with linear substitution |
| Difficulty | Moderate -0.8 This is a straightforward P3 integration question testing standard techniques: (i) is a simple substitution with a linear denominator requiring only the chain rule in reverse, and (ii) uses algebraic division followed by logarithmic integration. Both parts are routine textbook exercises with no problem-solving insight required, making this easier than average but not trivial due to the algebraic manipulation needed. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int\frac{12}{(2x-1)^2}dx = \frac{-6}{(2x-1)}\ (+c)\) | M1 A1 | M1: Achieves \(\frac{A}{(2x-1)}\) or equivalent e.g. \(A(2x-1)^{-1}\). A1: Achieves \(\frac{-6}{(2x-1)}\) or \(-6(2x-1)^{-1}\). No requirement for \(+c\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{4x+3}{x+2} = 4 - \frac{5}{x+2}\) | M1 A1 | M1: Uses division or suitable method to find either \(A\) or \(B\). Using identity \(4x+3=A(x+2)+B\). May be implied by \(A=4\) or \(B=-5\). A1: Correct values giving \(4-\frac{5}{x+2}\) or \(4+\frac{-5}{x+2}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int\frac{4x+3}{x+2}dx = \int 4-\frac{5}{x+2}dx = 4x - 5\ln | x+2 | \ (+c)\) |
| \(\int_{-8}^{-5}\frac{4x+3}{x+2}dx = (-20-5\ln 3)-(-32-5\ln 6)\) | dM1 | dM1: Substitutes \(-5\) and \(-8\) into integrated function containing \(\ln |
| \(= 12 + 5\ln 2\) | A1 | Accept equivalent exact answers e.g. \(12+\ln 32\). SC: If reaches \(12+5\ln 2\) via \((-20-5\ln(-3))-(-32-5\ln(-6))\) without modulus signs, withhold final A1. |
# Question 3:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\frac{12}{(2x-1)^2}dx = \frac{-6}{(2x-1)}\ (+c)$ | M1 A1 | M1: Achieves $\frac{A}{(2x-1)}$ or equivalent e.g. $A(2x-1)^{-1}$. A1: Achieves $\frac{-6}{(2x-1)}$ or $-6(2x-1)^{-1}$. No requirement for $+c$. |
**(2 marks)**
---
## Part (ii)(a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{4x+3}{x+2} = 4 - \frac{5}{x+2}$ | M1 A1 | M1: Uses division or suitable method to find either $A$ or $B$. Using identity $4x+3=A(x+2)+B$. May be implied by $A=4$ or $B=-5$. A1: Correct values giving $4-\frac{5}{x+2}$ or $4+\frac{-5}{x+2}$. |
**(2 marks)**
---
## Part (ii)(b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int\frac{4x+3}{x+2}dx = \int 4-\frac{5}{x+2}dx = 4x - 5\ln|x+2|\ (+c)$ | M1 A1ft | M1: Integrates $\frac{B}{x+2}\to ...\ln|x+2|$. Condone $\ln(x+2)$ or $\ln x+2$. A1ft: Integrates $A+\frac{B}{x+2}\to Ax+B\ln|x+2|$. Condone missing brackets unless implied by subsequent work. |
| $\int_{-8}^{-5}\frac{4x+3}{x+2}dx = (-20-5\ln 3)-(-32-5\ln 6)$ | dM1 | dM1: Substitutes $-5$ and $-8$ into integrated function containing $\ln|x+2|$ or $\ln(x+2)$. |
| $= 12 + 5\ln 2$ | A1 | Accept equivalent exact answers e.g. $12+\ln 32$. **SC:** If reaches $12+5\ln 2$ via $(-20-5\ln(-3))-(-32-5\ln(-6))$ without modulus signs, withhold final A1. |
**(4 marks) — Total: 8 marks**
---\begin{enumerate}
\item (i) Find
\end{enumerate}
$$\int \frac { 12 } { ( 2 x - 1 ) ^ { 2 } } \mathrm {~d} x$$
giving your answer in simplest form.\\
(ii) (a) Write $\frac { 4 x + 3 } { x + 2 }$ in the form
$$A + \frac { B } { x + 2 } \text { where } A \text { and } B \text { are constants to be found }$$
(b) Hence find, using algebraic integration, the exact value of
$$\int _ { - 8 } ^ { - 5 } \frac { 4 x + 3 } { x + 2 } d x$$
giving your answer in simplest form.\\
\hfill \mbox{\textit{Edexcel P3 2021 Q3 [8]}}