| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Integration with substitution and partial fractions |
| Difficulty | Standard +0.3 This is a structured multi-part question that guides students through a standard integration technique. Part (i) is routine substitution verification, part (ii) is textbook partial fractions with a repeated factor, and part (iii) combines the results. While it requires multiple techniques, each step is clearly signposted with no novel insight needed, making it slightly easier than average. |
| Spec | 1.02y Partial fractions: decompose rational functions1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(I = \int \frac{1}{u^2(1+u)^2} \times 2u\,du = \int \frac{2}{u(1+u)^2}\,du\) | M1 | For any attempt to find \(\frac{dx}{du}\) or \(\frac{du}{dx}\) |
| A1 | For '\(\mathbf{d}x = 2u\,\mathbf{d}u\)' or equivalent correctly used | |
| A1 | For showing the given result correctly | |
| 3 | ||
| (ii) \(2 = A(1+u)^2 + Bu(1+u) + Cu\) | M1 | For correct identity stated |
| \(A = 2\) | B1 | For correct value stated |
| \(C = -2\) | B1 | For correct value stated |
| \(0 = A + B\) (e.g.) | A1 | For any correct equation involving \(B\) |
| \(B = -2\) | A1 | For correct value |
| 5 | ||
| (iii) \(2\ln u - 2\ln(1+u) + \frac{2}{1+u}\) | B1 | For \(A\ln u + B\ln(1+u)\) with their values |
| B1 | For \(-C(1+u)^{-1}\) with their value | |
| Hence \(I = \ln x - 2\ln(1+\sqrt{x}) + \frac{2}{1+\sqrt{x}} + c\) | M1 | For substituting back |
| A1 | For completely correct answer (excluding \(c\)) | |
| 4 |
**(i)** $I = \int \frac{1}{u^2(1+u)^2} \times 2u\,du = \int \frac{2}{u(1+u)^2}\,du$ | M1 | For any attempt to find $\frac{dx}{du}$ or $\frac{du}{dx}$
| A1 | For '$\mathbf{d}x = 2u\,\mathbf{d}u$' or equivalent correctly used
| A1 | For showing the given result correctly
| **3** |
**(ii)** $2 = A(1+u)^2 + Bu(1+u) + Cu$ | M1 | For correct identity stated
$A = 2$ | B1 | For correct value stated
$C = -2$ | B1 | For correct value stated
$0 = A + B$ (e.g.) | A1 | For any correct equation involving $B$
$B = -2$ | A1 | For correct value
| **5** |
**(iii)** $2\ln u - 2\ln(1+u) + \frac{2}{1+u}$ | B1 | For $A\ln u + B\ln(1+u)$ with their values
| B1 | For $-C(1+u)^{-1}$ with their value
Hence $I = \ln x - 2\ln(1+\sqrt{x}) + \frac{2}{1+\sqrt{x}} + c$ | M1 | For substituting back
| A1 | For completely correct answer (excluding $c$)
| **4** |8 Let $I = \int \frac { 1 } { x ( 1 + \sqrt { } x ) ^ { 2 } } \mathrm {~d} x$.\\
(i) Show that the substitution $u = \sqrt { } x$ transforms $I$ to $\int \frac { 2 } { u ( 1 + u ) ^ { 2 } } \mathrm {~d} u$.\\
(ii) Express $\frac { 2 } { u ( 1 + u ) ^ { 2 } }$ in the form $\frac { A } { u } + \frac { B } { 1 + u } + \frac { C } { ( 1 + u ) ^ { 2 } }$.\\
(ii) Hence find $I$.
\hfill \mbox{\textit{OCR C4 Q8 [12]}}