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]