Find \(f'(x)\).

Hence find the \(x\)-coordinates of the points where the gradient of the graph of \(f\) is zero.

Find \(f''(x)\) expressing your answer in the form \(\frac{{p(x)}}{{{{({x^2} + 1)}^3}}}\), where \(p(x)\) is a polynomial of degree 3.

Find the \(x\)-coordinates of the other two points of inflexion.

Find the area of the shaded region. Express your answer in the form \(\frac{\pi }{a} - \ln \sqrt b \), where \(a\) and \(b\) are integers.

(a)     \(f'(x) = \frac{{\left( {{x^2} + 1} \right) - 2x(x + 1)}}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{ }}\left( { = \frac{{ - {x^2} - 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right)\)     M1A1

[2 marks]

\(\frac{{ - {x^2} - 2x + 1}}{{{{\left( {{x^2} + 1} \right)}^2}}} = 0\)

\(x =  - 1 \pm \sqrt 2 \)     A1

[1 mark]

\(f''(x) = \frac{{( - 2x - 2){{\left( {{x^2} + 1} \right)}^2} - 2(2x)\left( {{x^2} + 1} \right)\left( { - {x^2} - 2x + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^4}}}\)     A1A1

Note:     Award A1 for \(( - 2x - 2){\left( {{x^2} + 1} \right)^2}\) or equivalent.

Note:     Award A1 for \( - 2(2x)\left( {{x^2} + 1} \right)\left( { - {x^2} - 2x + 1} \right)\) or equivalent.

\( = \frac{{( - 2x - 2)\left( {{x^2} + 1} \right) - 4x\left( { - {x^2} - 2x + 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}}\)

\( = \frac{{2{x^3} + 6{x^2} - 6x - 2}}{{{{\left( {{x^2} + 1} \right)}^3}}}\)     A1

\(\left( { = \frac{{2\left( {{x^3} + 3{x^2} - 3x - 1} \right)}}{{{{\left( {{x^2} + 1} \right)}^3}}}} \right)\)

[3 marks]

recognition that \((x - 1)\) is a factor     (R1)

\((x - 1)\left( {{x^2} + bx + c} \right) = \left( {{x^3} + 3{x^2} - 3x - 1} \right)\)     M1

\( \Rightarrow {x^2} + 4x + 1 = 0\)     A1

\(x =  - 2 \pm \sqrt 3 \)     A1

Note:     Allow long division / synthetic division.

[4 marks]

\(\int_{ - 1}^0 {\frac{{x + 1}}{{{x^2} + 1}}{\text{d}}x} \)     M1

\(\int {\frac{{x + 1}}{{{x^2} + 1}}{\text{d}}x = \int {\frac{x}{{{x^2} + 1}}{\text{d}}x + \int {\frac{1}{{{x^2} + 1}}{\text{d}}x} } } \)     M1

\( = \frac{1}{2}\ln \left( {{x^2} + 1} \right) + \arctan (x)\)     A1A1

\( = \left[ {\frac{1}{2}\ln \left( {{x^2} + 1} \right) + \arctan (x)} \right]_{ - 1}^0 = \frac{1}{2}\ln 1 + \arctan 0 - \frac{1}{2}\ln 2 - \arctan ( - 1)\)     M1

\( = \frac{\pi }{4} - \ln \sqrt 2 \)     A1

[6 marks]