| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Improper integrals with partial fractions (infinite limit) |
| Difficulty | Challenging +1.2 This is a Further Maths Core Pure question requiring partial fractions decomposition, integration of rational functions (including arctangent), and evaluation of an improper integral. While it involves multiple techniques and careful limit evaluation, the structure is standard: decompose, integrate term-by-term, apply limits. The algebraic manipulation is moderate and the final form guides students to the answer. Harder than typical A-level integration but routine for Further Maths students who have practiced this topic. |
| Spec | 1.02y Partial fractions: decompose rational functions4.08c Improper integrals: infinite limits or discontinuous integrands4.08g Derivatives: inverse trig and hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{8x-12}{(2x^2+3)(x+1)} = \dfrac{Ax+B}{2x^2+3} + \dfrac{C}{x+1}\) | M1 | 3.1a - Selects correct form for partial fractions |
| \(8x-12 = (Ax+B)(x+1) + C(2x^2+3)\); e.g. \(x=-1 \Rightarrow C=-4\), \(x=0 \Rightarrow B=0\), \(x=1 \Rightarrow A=8\); or compares coefficients: \(A+2C=0\), \(A+B=8\), \(B+3C=-12\) | dM1 | 1.1b - Full method for all three constants; dependent on correct form |
| \(A=8,\ B=0,\ C=-4\) | A1 | 1.1b - Correct constants |
| \(\displaystyle\int\!\left(\frac{8x}{2x^2+3} - \frac{4}{x+1}\right)dx = 2\ln(2x^2+3) - 4\ln(x+1)\) | A1ft | 1.1b - Follow through on their constants |
| \(2\ln(2x^2+3)-4\ln(x+1) = \ln\!\left(\dfrac{(2x^2+3)^2}{(x+1)^4}\right)\) or \(= 2\ln\!\left(\dfrac{2x^2+3}{(x+1)^2}\right)\) | M1 | 2.1 - Combines logarithms correctly |
| \(\lim_{x\to\infty}\left\{\ln\dfrac{(2x^2+3)^2}{(x+1)^4}\right\} = \ln 4\) or \(\lim_{x\to\infty}\left\{2\ln\dfrac{2x^2+3}{(x+1)^2}\right\} = 2\ln 2\) | B1 | 2.2a |
| \(\displaystyle\int_0^{\infty} \frac{8x-12}{(2x^2+3)(x+1)}\,dx = \ln\dfrac{4}{9}\) | A1 | 1.1b - cao |
# Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{8x-12}{(2x^2+3)(x+1)} = \dfrac{Ax+B}{2x^2+3} + \dfrac{C}{x+1}$ | M1 | 3.1a - Selects correct form for partial fractions |
| $8x-12 = (Ax+B)(x+1) + C(2x^2+3)$; e.g. $x=-1 \Rightarrow C=-4$, $x=0 \Rightarrow B=0$, $x=1 \Rightarrow A=8$; or compares coefficients: $A+2C=0$, $A+B=8$, $B+3C=-12$ | dM1 | 1.1b - Full method for all three constants; dependent on correct form |
| $A=8,\ B=0,\ C=-4$ | A1 | 1.1b - Correct constants |
| $\displaystyle\int\!\left(\frac{8x}{2x^2+3} - \frac{4}{x+1}\right)dx = 2\ln(2x^2+3) - 4\ln(x+1)$ | A1ft | 1.1b - Follow through on their constants |
| $2\ln(2x^2+3)-4\ln(x+1) = \ln\!\left(\dfrac{(2x^2+3)^2}{(x+1)^4}\right)$ or $= 2\ln\!\left(\dfrac{2x^2+3}{(x+1)^2}\right)$ | M1 | 2.1 - Combines logarithms correctly |
| $\lim_{x\to\infty}\left\{\ln\dfrac{(2x^2+3)^2}{(x+1)^4}\right\} = \ln 4$ or $\lim_{x\to\infty}\left\{2\ln\dfrac{2x^2+3}{(x+1)^2}\right\} = 2\ln 2$ | B1 | 2.2a |
| $\displaystyle\int_0^{\infty} \frac{8x-12}{(2x^2+3)(x+1)}\,dx = \ln\dfrac{4}{9}$ | A1 | 1.1b - cao |\begin{enumerate}
\item Show that
\end{enumerate}
$$\int _ { 0 } ^ { \infty } \frac { 8 x - 12 } { \left( 2 x ^ { 2 } + 3 \right) ( x + 1 ) } \mathrm { d } x = \ln k$$
where $k$ is a rational number to be found.
\hfill \mbox{\textit{Edexcel CP1 2019 Q2 [7]}}