## Question 8:

### Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{-2x}{1-x^2}$ | B1 | Correct derivative |
| $1+\left(\frac{dy}{dx}\right)^2 = 1+\frac{4x^2}{(1-x^2)^2} = \frac{(1-x^2)^2+4x^2}{(1-x^2)^2}$ or $\frac{x^4-2x^2+1+4x^2}{(1-x^2)^2}$ or $\frac{x^4+2x^2+1}{(1-x^2)^2}$ | M1 | Attempts $1+\left(\frac{dy}{dx}\right)^2$, finds common denominator and shows working in numerator; condoning sign slips only (denominator may be expanded) |
| $= \frac{(1+x^2)^2}{(1-x^2)^2}$ or $\left(\frac{1+x^2}{1-x^2}\right)^2$ | A1 | Fully correct expression with factorised numerator and denominator |
| $\int_{\frac{1}{2}}^{\frac{3}{4}}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \int_{\frac{1}{2}}^{\frac{3}{4}}\left(\frac{1+x^2}{1-x^2}\right)dx$ | A1* | Fully correct proof with no errors; allow $x^2+1$ for $1+x^2$; allow $\int_{\frac{1}{2}}^{\frac{3}{4}}\frac{(1+x^2)}{(1-x^2)}dx$ or $\int_{\frac{1}{2}}^{\frac{3}{4}}\frac{1+x^2}{1-x^2}dx$ |

**(4 marks)**

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### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{(x^2+1)}{(1-x^2)} = -1+\frac{2}{1-x^2}$ or e.g. $-1+\frac{1}{1-x}+\frac{1}{1+x}$ | B1 | Writes the improper fraction correctly |
| $\int\frac{k}{1-x^2}\,dx = \pm\alpha\ln\frac{1+x}{1-x}$ **or** $\pm\alpha\ln(1+x)\pm\alpha\ln(1-x)$ **or** $\pm\alpha\,\text{artanh}\,x$ | M1 | Achieves an acceptable form for $\int\frac{k}{1-x^2}\,dx$ ($k$ constant); may see partial fraction approach |
| $\int-1+\frac{2}{1-x^2}\,dx = -x+\ln\frac{1+x}{1-x}$ | A1 | Correct integration |
| $\left[-x+\ln\frac{1+x}{1-x}\right]_{\frac{1}{2}}^{\frac{3}{4}} = -\frac{3}{4}+\ln 7-\left(-\frac{1}{2}+\ln 3\right)$ | dM1 | Evidence that the given limits have been applied; condone slips as long as intention is clear; **depends on the previous M** |
| $= -\frac{1}{4}+\ln\frac{7}{3}$ | A1 | cao |

**(5 marks)**

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**Note on common incorrect approach:**

$\int\frac{(1+x^2)}{(1-x^2)}\,dx = \int\left(\frac{1}{1-x^2}+\frac{x^2}{1-x^2}\right)dx = \frac{1}{2}\ln\frac{1+x}{1-x}+\ldots$

- If no attempt at $\int\left(\frac{x^2}{1-x^2}\right)dx$: scores B0M1A0M0A0
- If attempt at $\int\left(\frac{x^2}{1-x^2}\right)dx$ with limits applied: scores B0M1A0M1A0
- Partial fractions $\frac{1+x^2}{1-x^2}\equiv\frac{A}{1-x}+\frac{B}{1+x}$ generally score no marks
- $\frac{1+x^2}{1-x^2}\equiv\frac{A}{1-x}+\frac{B}{1+x}+C$ is correct and could score full marks

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**Alternative substitution approach for (b):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=\tanh\theta \Rightarrow \int\frac{(1+x^2)}{(1-x^2)}\,dx = \int\frac{(1+\tanh^2\theta)}{(1-\tanh^2\theta)}\text{sech}^2\theta\,d\theta$ | B1 | Substitutes fully |
| $\int\frac{(1+\tanh^2\theta)}{(1-\tanh^2\theta)}\text{sech}^2\theta\,d\theta = \int(1+\tanh^2\theta)\,d\theta = \int(2-\text{sech}^2\theta)\,d\theta$ | M1 | Cancel and applies $\tanh^2\theta = 1-\text{sech}^2\theta$ |
| $= \int(2-\text{sech}^2\theta)\,d\theta = 2\theta - \tanh\theta$ | A1 | Correct integration |
| $\left[2\,\text{artanh}\,x - x\right]_{\frac{1}{2}}^{\frac{3}{4}} = 2\times\frac{1}{2}\ln\left(\frac{1+\frac{3}{4}}{1-\frac{3}{4}}\right)-\frac{3}{4}-\left(2\times\frac{1}{2}\ln\left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right)-\frac{1}{2}\right)$ | dM1 | Evidence limits applied; condone slips; **depends on previous M** |
| $= -\frac{1}{4}+\ln\frac{7}{3}$ | A1 | cao |

**(5 marks)**

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