OCR Further Pure Core 1 2018 September — Question 7 6 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2018
SessionSeptember
Marks6
TopicIntegration with Partial Fractions
TypePartial fractions with irreducible quadratic
DifficultyChallenging +1.2 This is a Further Maths question requiring factorization of a cubic (which factors as (x-1)(x²+1)), partial fractions decomposition with an irreducible quadratic, and integration involving both ln and arctan. While it requires multiple techniques and careful algebra, it follows a standard template for this topic with no unusual tricks. The 'show detailed reasoning' instruction and specific answer form add modest difficulty, but this remains a typical Further Pure exam question rather than requiring novel insight.
Spec1.02y Partial fractions: decompose rational functions
7 In this question you must show detailed reasoning.
Find \(\int _ { 2 } ^ { 3 } \frac { x + 1 } { x ^ { 3 } - x ^ { 2 } + x - 1 } \mathrm {~d} x\), expressing your answer in the form \(a \ln b\) where \(a\) and \(b\) are rational numbers.
AnswerMarks Guidance
\(\frac{x+1}{x^3-x^2+x-1} = \frac{x+1}{(x^2+1)(x-1)}\)M1 Factorise denominator
\(= \frac{Ax+B}{(x^2+1)} + \frac{C}{(x-1)}\)M1 Split into partial fractions
\(\Rightarrow (Ax+B)(x-1)+C(x^2+1) = x+1\); \(x=1: 2C = 2 \Rightarrow C = 1\); \(x=0: -B+C = 1 \Rightarrow B = 0\); \(x=2: 2A+5 = 3 \Rightarrow A = -1\)M1 Valid method for finding \(A, B\) and \(C\)
\(\Rightarrow \frac{x+1}{(x^2+1)(x-1)} = \frac{1}{(x-1)} - \frac{x}{(x^2+1)}\)A1
\(\Rightarrow \int_2^3 \frac{x+1}{x^3-x^2+x-1}dx = \int_2^3(\frac{1}{(x-1)} - \frac{x}{(x^2+1)})dx\)M1 Substituting correct limits into their integral of the form \(\alpha\ln(x-1) + \beta\ln(x^2+1)\)
\(= [\ln(x-1) - \frac{1}{2}\ln(x^2+1)]_2^3 = \ln 2 - \frac{1}{2}\ln 10 + \frac{1}{2}\ln 5\)M1 Uses formula for integral
\(= \frac{1}{2}\ln 2\)A1 [6] oe in correct form (not eg ln√2) 
$\frac{x+1}{x^3-x^2+x-1} = \frac{x+1}{(x^2+1)(x-1)}$ | M1 | Factorise denominator

$= \frac{Ax+B}{(x^2+1)} + \frac{C}{(x-1)}$ | M1 | Split into partial fractions

$\Rightarrow (Ax+B)(x-1)+C(x^2+1) = x+1$; $x=1: 2C = 2 \Rightarrow C = 1$; $x=0: -B+C = 1 \Rightarrow B = 0$; $x=2: 2A+5 = 3 \Rightarrow A = -1$ | M1 | Valid method for finding $A, B$ and $C$

$\Rightarrow \frac{x+1}{(x^2+1)(x-1)} = \frac{1}{(x-1)} - \frac{x}{(x^2+1)}$ | A1 |

$\Rightarrow \int_2^3 \frac{x+1}{x^3-x^2+x-1}dx = \int_2^3(\frac{1}{(x-1)} - \frac{x}{(x^2+1)})dx$ | M1 | Substituting correct limits into their integral of the form $\alpha\ln(x-1) + \beta\ln(x^2+1)$

$= [\ln(x-1) - \frac{1}{2}\ln(x^2+1)]_2^3 = \ln 2 - \frac{1}{2}\ln 10 + \frac{1}{2}\ln 5$ | M1 | Uses formula for integral

$= \frac{1}{2}\ln 2$ | A1 [6] | oe in correct form (not eg ln√2)

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7 In this question you must show detailed reasoning.\\
Find $\int _ { 2 } ^ { 3 } \frac { x + 1 } { x ^ { 3 } - x ^ { 2 } + x - 1 } \mathrm {~d} x$, expressing your answer in the form $a \ln b$ where $a$ and $b$ are rational numbers.

\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q7 [6]}}
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