# Question 5:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\{x = u^2 \Rightarrow\right\} \frac{dx}{du} = 2u$ or $\frac{du}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$ or $\frac{du}{dx} = \frac{1}{2\sqrt{x}}$ | B1 | Also accept $dx = 2u\, du$ or $du = \frac{dx}{2\sqrt{x}}$ |
| $\left\{\int \frac{1}{x(2\sqrt{x}-1)}dx\right\} = \int \frac{1}{u^2(2u-1)} \cdot 2u\, du$ | M1 | Full substitution producing integral in $u$ only (including $du$); must deal with $x$, $(2\sqrt{x}-1)$ and $dx$ |
| $= \int \frac{2}{u(2u-1)}\, du$ | A1* cso | Leading to printed result (including $du$) |

**[3 marks]**

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2}{u(2u-1)} \equiv \frac{A}{u} + \frac{B}{(2u-1)} \Rightarrow 2 \equiv A(2u-1) + Bu$ | M1 A1 | Writing partial fractions form and finding at least one of $A$ or $B$; both $A=-2$ and $B=4$ |
| $u=0 \Rightarrow A=-2$; $u=\frac{1}{2} \Rightarrow B=4$ | | |
| $\int \frac{2}{u(2u-1)}\,du = \int \frac{-2}{u} + \frac{4}{(2u-1)}\,du$ | M1 | Integrates $\frac{M}{u} + \frac{N}{(2u-1)}$, $M\neq 0$, $N\neq 0$ to obtain $\pm\lambda\ln u$ or $\pm\mu\ln(2u-1)$ |
| $= -2\ln u + 2\ln(2u-1)$ | A1 ft | At least one term correctly followed through |
| | A1 cao | $-2\ln u + 2\ln(2u-1)$ |
| $\left[-2\ln u + 2\ln(2u-1)\right]_1^3$ | M1 | Applies limits of 3 and 1 in $u$ (or 9 and 1 in $x$) and subtracts correct way round |
| $= (-2\ln 3 + 2\ln 5) - (-2\ln 1 + 2\ln 1) = -2\ln 3 + 2\ln 5 - 0$ | | |
| $= 2\ln\left(\frac{5}{3}\right)$ | A1 cso cao | Correct answer only; note $a=5$, $b=3$ |

**Important notes:**
- Award M0A0M1A1ftA0 M1A0 for $\int\frac{2}{u(2u-1)}du = \int\frac{2}{u}+\frac{2}{(2u-1)}du = 2\ln u + \ln(2u-1)$ (evidence partial fractions given)
- Award M0A0M0A0A0 M1A0 for writing $2\ln u + 2\ln(2u-1)$ or $2\ln u + \ln(2u-1)$ without evidence of partial fractions
- Award M1A1M1A1A1 for $-2\ln u + 2\ln(2u-1)$ without evidence of partial fractions

**[7 marks]**