$\frac{dy}{dx} = \frac{y}{x(x+1)}$ | M1 | correctly separating variables and intending to integrate (ie need to see attempt at integration or integral signs)

$\frac{1}{y}dy = \frac{1}{x(x+1)}dx$ | M1 | partial fractions soi

$\frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1}$ | A1 | $A = 1$ www

$1 = A(x+1) + Bx$ | A1 | $B = -1$ www

$x = 0 \Rightarrow A = 1$ | A1 | ft their $A,B$

$x = 1 \Rightarrow 1 = B \Rightarrow B = -1$ | B1 | condone absence of $c$ or $\ln c$

$\ln y = \left(\frac{1}{x} - \frac{1}{x+1}\right)dx = \ln x - \ln(x+1) + c$ | M1 | evaluating their $c$ at any stage dependent on $x$ and $y$ terms all being logs of correct form but do not award following incorrect log rules, ft their $A,B$. $c$ could be say a decimal. (eg $y = \frac{x}{x+1} + c$ then $c$ being found is B0)

$x = 1, y = 1 \Rightarrow 0 = 0 - \ln 2 + c \Rightarrow c = \ln 2$ | A1 | correctly combining lns and antilogging throughout (must have included the constant term). Apply this strictly. Do not allow if $c$ is included as an afterthought unless completely convinced. ft $A,B$. Logs must be of correct form ie not following $\int \frac{1}{x(x+1)}dx = \ln(x^2-x)$ unless ft from partial fractions and $B = -1$

$\ln y = \ln x - \ln(x+1) + \ln 2 = \ln\left(\frac{2x}{x+1}\right)$ | | 

$y = \frac{2x}{x+1}$ | A1 | cao www ($y = e^{0.693}$ loses final A1)

**Note:** evaluating $c$ and log work can be in either order. eg $y = \frac{cx}{x+1}$, at $x = 1, y = 1$, $c = 2$

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