Use the quotient rule to show that $f'(x) = \frac{{10 - 2{x^2}}}{{{{({x^2} + 5)}^2}}}$.

Find $\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x}$.

The following diagram shows part of the graph of $f$.

The shaded region is enclosed by the graph of $f$, the $x$-axis, and the lines $x = \sqrt 5$ and $x = q$. This region has an area of $\ln 7$. Find the value of $q$.

derivative of $2x$ is $2$ (must be seen in quotient rule)     (A1)

derivative of ${x^2} + 5$ is $2x$ (must be seen in quotient rule)     (A1)

correct substitution into quotient rule     A1

eg     $\frac{{({x^2} + 5)(2) - (2x)(2x)}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2({x^2} + 5) - 4{x^2}}}{{{{({x^2} + 5)}^2}}}$

correct working which clearly leads to given answer   A1

eg    $\frac{{2{x^2} + 10 - 4{x^2}}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2{x^2} + 10 - 4{x^2}}}{{{x^4} + 10{x^2} + 25}}$

$f'(x) = \frac{{10 - 2{x^2}}}{{{{({x^2} + 5)}^2}}}$     AG     N0

[4 marks]

valid approach using substitution or inspection     (M1)

eg     $u = {x^2} + 5,{\text{ d}}u = 2x{\text{d}}x,{\text{ }}\frac{1}{2}\ln ({x^2} + 5)$

$\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x = \int {\frac{1}{u}{\text{d}}u} }$     (A1)

$\int {\frac{1}{u}{\text{d}}u = \ln u + c}$     (A1)

$\ln ({x^2} + 5) + c$     A1     N4

[4 marks]

correct expression for area     (A1)

eg     $\left[ {\ln \left( {{x^2} + 5} \right)} \right]_{\sqrt 5 }^q,{\text{ }}\int\limits_{\sqrt 5 }^q {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x}$

substituting limits into their integrated function and subtracting (in either order)     (M1)

eg     $\ln ({q^2} + 5) - \ln \left( {{{\sqrt 5 }^2} + 5} \right)$

correct working     (A1)

eg     $\ln \left( {{q^2} + 5} \right) - \ln 10,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}}$

equating their expression to $\ln 7$ (seen anywhere)     (M1)

eg     $\ln \left( {{q^2} + 5} \right) - \ln 10 = \ln 7,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}} = \ln 7,{\text{ }}\ln ({q^2} + 5) = \ln 7 + \ln 10$

correct equation without logs     (A1)

eg     $\frac{{{q^2} + 5}}{{10}} = 7,{\text{ }}{q^2} + 5 = 70$

${q^2} = 65$     (A1)

$q = \sqrt {65}$     A1     N3

Note: Award A0 for $q =  \pm \sqrt {65}$.

[7 marks]