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]