# Question 1:

## Part (a)(i)

**Answer** | **Mark** | **Guidance**

$\frac{(2x+1)\times 2x - x^2 \times 2}{(2x+1)^2}$ oe | B1 | $2x(2x+1)$ or $-2x^2$ in numerator; condone missing brackets

Correct denominator $(2x+1)^2$ | B1 | Allow correct equivalent forms

Correct numerator | B1 | ISW for further simplifications

e.g. $\frac{2x^2+2x}{(2x+1)^2}$ or $\frac{2x(x+1)}{(2x+1)^2}$ oe | | No need to see this explicitly

**Alternative method:** $x^2(-2)(2x+1)^{-2} + 2x(2x+1)^{-1}$ oe | B1 B1 B1 | $\pm 2x^2(2x+1)^{-2}$ oe; $+2x(2x+1)^{-1}$ oe; all correct. ISW for further simplifications

**[3]**

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## Part (a)(ii)

$(2x-3)\sec^2(x^2-3x)$ oe | B1 | B1 for $\sec^2(x^2-3x)$

| B1 | B1 for all correct; condone $\sec^2(x^2-3x)(2x-3)$; ISW for further simplifications

**[2]**

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## Part (b)

*Note: Allow without mod in both parts (b) and (c)*

$x=(u+1)^2,\ \frac{dx}{du}=2(u+1)$ oe | M1 | EITHER attempt $x$ in terms of $u$ and differentiate, OR attempt $\frac{du}{dx}$ and obtain $kx^{-0.5}$ oe; allow in form $dx=\ldots$ or $du=\ldots$

or $\frac{du}{dx}=0.5x^{-0.5}$ oe | |

$2\int\frac{(u+1)}{u}\,du$ or $2\int\!\left(1+\frac{1}{u}\right)du$ oe | A1 | Allow $k\int\frac{(u+1)}{u}\,du$ or $k\int\!\left(1+\frac{1}{u}\right)du$; or $\int\frac{(ku+j)}{u}\,du$ or $\int\!\left(k+\frac{j}{u}\right)du$

$=2(u+\ln|u|)$ $(+c)$ | A1 | Allow without $+c$ here

$=2(\sqrt{x}-1+\ln|\sqrt{x}-1|)+c$ oe | A1 | All correct incl $+c$; **not penalise $+c$ in both (b) & (c)**; ISW for further simplifications

or $2(\sqrt{x}+\ln|\sqrt{x}-1|)+c$ oe | | Integration by parts: use same scheme.

or $2\sqrt{x}+\ln(\sqrt{x}-1)^2+c$ oe | |

**[4]**

# Question 1:

## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln|2x^2 - 8x - 1|$ or $\ln|\frac{1}{2}x^2 - 2x - \frac{1}{4}|$ seen | M1 | or $u = 2x^2 - 8x - 1$ and $\ln|u|$ seen; or $u = x-2$ and $\ln|2u^2 - 9|$ seen |
| $\frac{1}{4}\ln|2x^2 - 8x - 1| + c$ or $\frac{1}{4}\ln|\frac{1}{2}x^2 - 2x - \frac{1}{4}| + c$ | A1 | All correct including $+c$; Correct answer seen: M1A1 even if eg $(x-2)\frac{\ln|2x^2-8x-1|}{4x-8} = \frac{1}{4}\ln|2x^2-8x-1|$; Not penalise $+c$ in both (b) & (c); ISW for further simplifications |

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