| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Partial fractions with irreducible quadratic |
| Difficulty | Standard +0.8 This is a Further Maths Core Pure question requiring partial fractions with an irreducible quadratic, then integration involving both logarithmic and arctangent forms. While the partial fractions setup is standard, students must correctly handle the quadratic term (requiring Bx+C in the numerator), integrate arctan(x/2), and manipulate the final expression into the required form with ln(a√2). This combines multiple techniques and requires careful algebraic manipulation, making it moderately challenging but still within expected Further Maths scope. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08j Integration using partial fractions4.05c Partial fractions: extended to quadratic denominators |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{2x^2+3x+6}{(x+1)(x^2+4)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+4} \Rightarrow 2x^2+3x+6 = A(x^2+4)+(Bx+C)(x+1)\) | M1 | Selects correct form for partial fractions and multiplies through to form suitable identity, or uses method to find at least one value. |
| e.g. \(x=-1 \Rightarrow A=\ldots\), \(x=0 \Rightarrow C=\ldots\), coeff \(x^2 \Rightarrow B=\ldots\) or compares coefficients: \(2=A+B\), \(3=B+C\), \(6=4A+C\) | dM1 | Full method for finding values for all three constants. Dependent on first M. Allow slips as long as intention is clear. |
| \(A=1,\quad B=1,\quad C=2\) | A1 | Correct constants or partial fractions. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int_0^2 \frac{1}{x+1} + \frac{x}{x^2+4} + \frac{2}{x^2+4}\, dx = \left[\alpha\ln(x+1)+\beta\ln(x^2+4)+\lambda\arctan\!\left(\frac{x}{2}\right)\right]_0^2\) | M1 | Splits integral into integrable form and integrates at least two terms to correct form. May use substitution on arctan term. |
| \(= \left[\ln(x+1) + \frac{1}{2}\ln(x^2+4) + \arctan\!\left(\frac{x}{2}\right)\right]_0^2\) | A1 | Fully correct integration. |
| \(= \left[\ln(3) + \frac{1}{2}\ln(8) + \arctan(1)\right] - \left[\ln(1) + \frac{1}{2}\ln(4) + \arctan(0)\right]\) | dM1 | Uses limits 0 and 2, subtracts correct way round and combines ln terms from separate integrals to single term with evidence of correct ln laws at least once. |
| \(\ln(3\sqrt{2}) + \dfrac{\pi}{4}\) | A1 | Correct answer. |
## Question 6:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{2x^2+3x+6}{(x+1)(x^2+4)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+4} \Rightarrow 2x^2+3x+6 = A(x^2+4)+(Bx+C)(x+1)$ | M1 | Selects correct form for partial fractions and multiplies through to form suitable identity, or uses method to find at least one value. |
| e.g. $x=-1 \Rightarrow A=\ldots$, $x=0 \Rightarrow C=\ldots$, coeff $x^2 \Rightarrow B=\ldots$ or compares coefficients: $2=A+B$, $3=B+C$, $6=4A+C$ | dM1 | Full method for finding values for all three constants. Dependent on first M. Allow slips as long as intention is clear. |
| $A=1,\quad B=1,\quad C=2$ | A1 | Correct constants or partial fractions. |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_0^2 \frac{1}{x+1} + \frac{x}{x^2+4} + \frac{2}{x^2+4}\, dx = \left[\alpha\ln(x+1)+\beta\ln(x^2+4)+\lambda\arctan\!\left(\frac{x}{2}\right)\right]_0^2$ | M1 | Splits integral into integrable form and integrates at least two terms to correct form. May use substitution on arctan term. |
| $= \left[\ln(x+1) + \frac{1}{2}\ln(x^2+4) + \arctan\!\left(\frac{x}{2}\right)\right]_0^2$ | A1 | Fully correct integration. |
| $= \left[\ln(3) + \frac{1}{2}\ln(8) + \arctan(1)\right] - \left[\ln(1) + \frac{1}{2}\ln(4) + \arctan(0)\right]$ | dM1 | Uses limits 0 and 2, subtracts correct way round and combines ln terms from separate integrals to single term with evidence of correct ln laws at least once. |
| $\ln(3\sqrt{2}) + \dfrac{\pi}{4}$ | A1 | Correct answer. |
---\begin{enumerate}
\item (a) Express as partial fractions
\end{enumerate}
$$\frac { 2 x ^ { 2 } + 3 x + 6 } { ( x + 1 ) \left( x ^ { 2 } + 4 \right) }$$
(b) Hence, show that
$$\int _ { 0 } ^ { 2 } \frac { 2 x ^ { 2 } + 3 x + 6 } { ( x + 1 ) \left( x ^ { 2 } + 4 \right) } d x = \ln ( a \sqrt { 2 } ) + b \pi$$
where $a$ and $b$ are constants to be determined.
\hfill \mbox{\textit{Edexcel CP1 2022 Q6 [7]}}