(a)

\begin{align} \int x^2(x - 3) \phantom{.} dx & = \int x^3 - 3x^2 \phantom{.} dx \\ & = {x^4 \over 4} - {3x^3 \over 3} \\ & = {1 \over 4}x^4 - x^3 + c \end{align}

(b)

\begin{align} \int {2x^2 - \sqrt{x} \over x} \phantom{.} dx & = \int {2x^2 \over x} - {x^{1 \over 2} \over x} \phantom{.} dx \\ & = \int 2x - x^{-{1 \over 2}} \phantom{.} dx \\ & = {2x^2 \over 2} - {x^{1 \over 2} \over {1 \over 2}} \\ & = x^2 - 2x^{1 \over 2} \\ & = x^2 - 2\sqrt{x} + c \end{align}

(a)

\begin{align} \int (2 + e^x)^2 \phantom{.} dx & = \int (2)^2 + 2(2)(e^x) + (e^x)^2 \phantom{.} dx \phantom{000000} [\text{Identity: } (a + b)^2 = a^2 + 2ab + b^2] \\ & = \int 4 + 4e^x + e^{2x} \phantom{.} dx \\ & = {4x \over 1} + {4e^x \over 1} + {e^{2x} \over 2} \\ & = 4x + 4e^x + {1 \over 2}e^{2x} + c \end{align}

(b)

\begin{align} \int {e^{3x} - 2 \over e^x} \phantom{.} dx & = \int {e^{3x} \over e^x} - {2 \over e^x} \phantom{.} dx \\ & = \int e^{2x} - 2e^{-x} \phantom{.} dx \\ & = {e^{2x} \over 2} - {2e^{-x} \over -1} \\ & = {1 \over 2}e^{2x} + 2e^{-x} \\ & = {1 \over 2}e^{2x} + {2 \over e^x} + c \end{align}

(c) The special values used are: $\sin {\pi \over 4} = \sin 45^\circ = {1 \over \sqrt{2}}$ and $\cos {\pi \over 4} = \cos 45^\circ = {1 \over \sqrt{2}}$

\begin{align} \int_0^{\pi \over 4} (\cos x - 3\sin x) \phantom{0} dx & = \left[ {\sin x \over 1} - {3 (-\cos x) \over 1} \right]_0^{\pi \over 4} \\ & = \left[ \sin x + 3\cos x \right]_0^{\pi \over 4} \\ & = \sin {\pi \over 4} + 3\cos {\pi \over 4} - \left[ \sin 0 + 3\cos 0 \right] \\ & = {1 \over \sqrt{2}} + 3 \left( {1 \over \sqrt{2}} \right) - \left[ 0 + 3(1) \right] \\ & = {1 \over \sqrt{2}} + {3 \over \sqrt{2}} - 3 \\ & = {4 \over \sqrt{2}} - 3 \\ & = {4 \over \sqrt{2}} \times {\sqrt{2} \over \sqrt{2}} - 3 \phantom{00000000} [\text{Rationalise denominator}] \\ & = {4\sqrt{2} \over 2} - 3 \\ & = 2\sqrt{2} - 3 \end{align}

(d) The special values used are: $\cos {\pi \over 3} = \cos 60^\circ = {1 \over 2}$ and $\cos {\pi \over 6} = \cos 30^\circ = {\sqrt{3} \over 2}$

\begin{align} \int_{\pi \over 4}^{\pi \over 3} \sin \left(2x - {\pi \over 3} \right) \phantom{.} dx & = \left[ {-\cos \left(2x - {\pi \over 3} \right) \over 2} \right]_{\pi \over 4}^{\pi \over 3} \\ & = {-\cos \left[2({\pi \over 3}) - {\pi \over 3}\right] \over 2} - \left[ {-\cos \left(2({\pi \over 4}) - {\pi \over 3}\right) \over 2} \right] \\ & = {-\cos {\pi \over 3} \over 2} - \left( {-\cos {\pi \over 6} \over 2} \right) \\ & = {- {1 \over 2} \over 2} - \left( {-{\sqrt{3} \over 2} \over 2} \right) \\ & = -{1 \over 4} + {\sqrt{3} \over 4} \\ & = {1 \over 4} \left(-1 + \sqrt{3} \right) \\ & = {1 \over 4} \left(\sqrt{3} - 1\right) \end{align}

(i)

\begin{align} u & = 1 + \cos x &&& v & = \sin x \\ {du \over dx} & = - \sin x &&& {dv \over dx} & = \cos x \end{align} \begin{align} \require{cancel} {d \over dx} \left(1 + \cos x \over \sin x \right) & = { (\sin x)(- \sin x) - (1 + \cos x)(\cos x) \over (\sin x)^2} \phantom{000000} [\text{Quotient rule}] \\ & = { - \sin^2 x - \cos x (1 + \cos x) \over \sin^2 x} \\ & = { - \sin^2 x - \cos x - \cos^2 x \over \sin^2 x} \\ & = { - \sin^2 x - \cos^2 x - \cos x \over \sin^2 x} \\ & = { - (\sin^2 x + \cos^2 x) - \cos x \over \sin^2 x} \\ & = { - 1 - \cos x \over 1 - \cos^2 x } \phantom{0000000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = { - \cancel{(1 + \cos x)} \over \cancel{(1 + \cos x)}(1 - \cos x) } \phantom{00000000} [a^2 - b^2 = (a + b)(a - b)] \\ & = { - 1 \over 1 - \cos x} \phantom{00} \text{ (Shown)} \end{align}

(ii) The special values used are $\cos {\pi \over 4} = \cos 45^\circ = {1 \over \sqrt{2}}$ and $\cos {\pi \over 4} = \cos 45^\circ = {1 \over \sqrt{2}}$

\begin{align} {d \over dx} \left(1 + \cos x \over \sin x \right) & = -{1 \over 1 - \cos x} \\ \\ \implies \int -{1 \over 1 - \cos x} \phantom{.} dx & = {1 + \cos x \over \sin x} \\ \\ \therefore \int_{\pi \over 4}^{\pi \over 2} {1 \over 1 - \cos x} \phantom{.} dx & = - \int_{\pi \over 4}^{\pi \over 2} - {1 \over 1 - \cos x} \phantom{.} dx \\ & = -\left[ {1 + \cos x \over \sin x} \right]_{\pi \over 4}^{\pi \over 2} \\ & = - \left[ {1 + \cos {\pi \over 2} \over \sin {\pi \over 2}} - {1 + \cos {\pi \over 4} \over \sin {\pi \over 4} } \right] \\ & = - \left[ {1 + 0 \over 1} - {1 + {1 \over \sqrt{2}} \over {1 \over \sqrt{2}}} \right] \\ & = - \left[ 1 - \sqrt{2} \left(1 + {1 \over \sqrt{2}} \right) \right] \\ & = - \left( 1 - \sqrt{2} - 1 \right) \\ & = - (- \sqrt{2}) \\ & = \sqrt{2} \end{align}

(i)

\begin{align} {1 \over (x - 3)(2x + 1)} & = {A \over x - 3} + {B \over 2x + 1} \\ {1 \over (x - 3)(2x + 1)} & = {A(2x + 1) \over (x - 3)(2x + 1)} + {B(x - 3) \over (x - 3)(2x + 1)} \\ {1 \over (x - 3)(2x + 1)} & = {A(2x + 1) + B(x - 3) \over (x - 3)(2x + 1)} \\ \\ \therefore 1 & = A(2x + 1) + B(x - 3) \\ \\ \text{Let } & x = -{1 \over 2}, \\ 1 & = A(0) + B\left(-{1 \over 2} - 3\right) \\ 1 & = 0 + B\left(-{7 \over 2}\right) \\ 1 & = -{7 \over 2}B \\ 2 & = -7B \\ -{2 \over 7} & = B \\ \\ \text{Let } & x = 3, \\ 1 & = A[2(3) + 1] + B(0) \\ 1 & = A(7) + 0 \\ 1 & = 7A \\ {1 \over 7} & = A \\ \\ \therefore {1 \over (x - 3)(2x + 1)} & = {{1 \over 7} \over x - 3} + {-{2 \over 7} \over 2x + 1} \\ & = {1 \over 7(x - 3)} - {2 \over 7(2x + 1)} \end{align}

(ii)

\begin{align} \int_4^7 {1 \over (x - 3)(2x + 1)} \phantom{.} dx & = \int_4^7 {1 \over 7(x - 3)} - {2 \over 7(2x + 1)} \phantom{.} dx \\ & = \int_4^7 {1 \over 7}\left(1 \over x - 3\right) - {2 \over 7}\left(1 \over 2x + 1\right) \phantom{.} dx \\ & = \left[ {1 \over 7}\left( \ln (x - 3) \over 1 \right) - {2 \over 7} \left( \ln (2x + 1) \over 2 \right) \right]_4^7 \\ & = \left[ {1 \over 7}\ln (x - 3) - {1 \over 7}\ln (2x + 1) \right]_4^7 \\ & = {1 \over 7} \left[ \ln (x - 3) - \ln (2x + 1) \right]_4^7 \\ & = {1 \over 7} \left[ \ln \left( {x - 3 \over 2x + 1}\right) \right]_4^7 \phantom{000000000000000000000} [\text{Quotient law (logarithms)}] \\ & = {1 \over 7} \left[ \ln \left( {7 - 3 \over 2(7) + 1} \right) - \ln \left({4 - 3 \over 2(4) + 1}\right) \right] \\ & = {1 \over 7} \left[ \ln {4 \over 15} - \ln {1 \over 9} \right] \\ & = {1 \over 7} \left[ \ln {{4 \over 15} \over {1 \over 9}} \right] \phantom{00000000000000000000000000} [\text{Quotient law (logarithms)}] \\ & = {1 \over 7} \ln {12 \over 5} \end{align}

(a) Note: $\cos {\pi \over 3} = \cos 60^\circ = {1 \over 2}$

\begin{align} {d \over dx} \left[ \ln (\cos x) \right] & = {-\sin x \over \cos x} \\ & = -\tan x \phantom{00} \text{ (Shown)} \\ \\ \\ \implies \int - \tan x \phantom{.} dx & = \ln (\cos x) \\ \\ \int_0^{\pi \over 3} \tan x \phantom{.} dx & = - \int_0^{\pi \over 3} - \tan x \phantom{.} dx \\ & = - [ \ln (\cos x) ]_0^{\pi \over 3} \\ & = - \left[ \ln \left(\cos {\pi \over 3} \right) - \ln (\cos 0) \right] \\ & = - \left( \ln {1 \over 2} - \ln 1 \right) \\ & = - \left( \ln {1 \over 2} - 0 \right) \\ & = - \ln {1 \over 2} \\ & = - \ln 2^{-1} \\ & = - (-1) \ln 2 \phantom{0000000} [\text{Power law (logarithms)}] \\ & = \ln 2 \end{align}

(b)

\begin{align} {d \over dx} \left[ \ln (e^x + 1) \right] & = {(1)e^x \over e^x + 1} \\ & = {e^x \over e^x + 1} \phantom{00} \text{ (Shown)} \\ \\ \\ \implies \int {e^x \over e^x + 1} \phantom{.} dx & = \ln (e^x + 1) \\ \\ \therefore \int_0^{\ln 2} {e^x \over e^x + 1} \phantom{.} dx & = \left[ \ln (e^x + 1) \right]_0^{\ln 2} \\ & = \ln (e^{\ln 2} + 1) - \ln (e^0 + 1) \\ & = \ln (2 + 1) - \ln (1 + 1) \\ & = \ln 3 - \ln 2 \\ & = \ln {3 \over 2} \phantom{00000000} [\text{Quotient law (logarithms)}] \end{align}

(a)

\begin{align} \int_1^a (3 - 2x) \phantom{.} dx & = \left[ {3x \over 1} - {2x^2 \over 2} \right]_1^a \\ & = \left[3x - x^2 \right]_1^a \\ & = \left[3(a) - (a)^2\right] - \left[3(1) - (1)^2\right] \\ & = (3a - a^2) - (3 - 1) \\ & = 3a - a^2 - 2 \\ \\ \int_2^1 4x \phantom{.} dx & = \left[ {4x^2 \over 2} \right]_2^1 \\ & = \left[2x^2\right]_2^1 \\ & = 2(1)^2 - 2(2)^2 \\ & = 2 - 8 \\ & = -6 \\ \\ \text{Since } \int_1^a (3 - 2x) & \phantom{.} dx = \int_2^1 4x \phantom{.} dx, \\ 3a - a^2 - 2 & = -6 \\ -a^2 + 3a + 4 & = 0 \\ a^2 - 3a - 4 & = 0 \\ (a - 4)(a + 1) & = 0 \\ \\ a - 4 = 0 \phantom{000} & \text{or} \phantom{000} a + 1 = 0 \\ a = 4 \phantom{000} & \phantom{or000+1} a = - 1 \end{align}

(b)

(i)

\begin{align} {dy \over dx} & = 3x^2 + k \\ \\ \text{When } & x = -2 \text{ and } {dy \over dx} = 0, \phantom{00000} [\text{T.p. } (-2, 6)] \\ 0 & = 3(-2)^2 + k \\ 0 & = 3(4) + k \\ 0 & = 12 + k \\ -12 & = k \end{align}

(ii)

\begin{align} y & = \int {dy \over dx} \phantom{.} dx \\ y & = \int (3x^2 - 12) \phantom{.} dx \\ y & = {3x^3 \over 3} - {12x \over 1} + c \\ y & = x^3 - 12x + c \\ \\ \text{Using } & (-2, 6), \\ 6 & = (-2)^3 - 12(-2) + c \\ 6 & = -8 + 24 + c \\ 6 + 8 - 24 & = c \\ -10 & = c \\ \\ \text{Eqn of curve: } & y = x^3 - 12x - 10 \\ \\ \\ \text{When } & x = 0, \\ y & = (0)^3 - 12(0) - 10 \\ & = -10 \\ \\ \therefore & \phantom{.} (0, -10) \end{align}

(a)

\begin{align} \int_0^4 \sqrt{2x + 1} \phantom{.} dx & = \int_0^4 (2x + 1)^{1 \over 2} \phantom{.} dx \\ & = \left[ { (2x + 1)^{3 \over 2} \over \left(3 \over 2\right)(2)} \right]_0^4 \\ & = \left[ { (2x + 1)^{3 \over 2} \over 3 } \right]_0^4 \\ & = {[2(4) + 1]^{3 \over 2} \over 3} - {[2(0) + 1]^{3 \over 2} \over 3} \\ & = 9 - {1 \over 3} \\ & = 8{2 \over 3} \end{align}

(b)

\begin{align} \int_1^4 (2x - \sqrt{x})^2 \phantom{.} dx & = \int_1^4 (2x)^2 - 2(2x)(\sqrt{x}) + (\sqrt{x})^2 \phantom{.} dx \phantom{000000} [(a - b)^2 = a^2 - 2ab + b^2] \\ & = \int_1^4 4x^2 - 4x^{3 \over 2} + x \phantom{.} dx \\ & = \left[ {4x^3 \over 3} - {4x^{5 \over 2} \over {5 \over 2}} + {x^2 \over 2} \right]_1^4 \\ & = \left[ {4 \over 3}x^3 - {8 \over 5} x^{5 \over 2} + {1 \over 2}x^2 \right]_1^4 \\ & = {4 \over 3}(4)^3 - {8 \over 5}(4)^{5 \over 2} + {1 \over 2}(4)^2 - \left[ {4 \over 3}(1)^3 - {8 \over 5}(1)^{5 \over 2} + {1 \over 2}(1)^2 \right] \\ & = 42{2 \over 15} - {7 \over 30} \\ & = 41{9 \over 10} \end{align}

(c)

\begin{align} \int_0^1 {(4x + 1)^4 - 7 \over 2(4x + 1)^2} \phantom{.} dx & = \int_0^1 {(4x + 1)^4 \over 2(4x + 1)^2} - {7 \over 2(4x + 1)^2} \phantom{.} dx \\ & = \int_0^1 {(4x + 1)^2 \over 2} - {7 \over 2}(4x + 1)^{-2} \phantom{.} dx \\ & = \int_0^1 {1 \over 2}(4x + 1)^2 - {7 \over 2}(4x + 1)^{-2} \phantom{.} dx \\ & = \left[ {1 \over 2} \left( (4x + 1)^3 \over (3)(4) \right) - {7 \over 2} \left( (4x + 1)^{-1} \over (-1)(4) \right) \right]_0^1 \\ & = \left[ {1 \over 24} (4x + 1)^3 + {7 \over 8} (4x + 1)^{-1} \right]_0^1 \\ & = {1 \over 24} [4(1) + 1]^3 + {7 \over 8} [4(1) + 1]^{-1} - \left\{ {1 \over 24} [4(0) + 1]^3 + {7 \over 8} [4(0) + 1]^{-1} \right\} \\ & = 5{23 \over 60} - {11 \over 12} \\ & = 4{7 \over 15} \end{align}

(a)

\begin{align} \int_0^2 {4 \over 2x + 1} \phantom{.} dx & = \left[ 4\ln (2x + 1) \over 2 \right]_0^2 \\ & = \left[2\ln(2x + 1) \right]_0^2 \\ & = 2\ln [2(2) + 1] - 2\ln [2(0) + 1] \\ & = 2\ln 5 - 2\ln 1 \\ & = 2\ln 5 - 2(0) \\ & = 2\ln 5 \end{align}

(b)

\begin{align} \int_1^3 {3 \over 5 - x} \phantom{.} dx & = \left[ 3\ln (5 - x) \over - 1 \right]_1^3 \\ & = \left[ -3\ln (5 - x) \right]_1^3 \\ & = \left[ - 3\ln (5 - 3) \right] - \left[ -3 \ln (5 - 1) \right] \\ & = \left[ - 3\ln 2 \right] - \left[ -3 \ln 4 \right] \\ & = -3 \ln 2 + 3\ln 4 \\ & = \ln 2^{-3} + \ln 4^3 \phantom{0000000000} [\text{Power law (logarithms)}] \\ & = \ln {1 \over 8} + \ln 64 \\ & = \ln \left[{1 \over 8}(64)\right] \phantom{000000000000} [\text{Product law (logarithms)}] \\ & = \ln 8 \end{align}

(c) Note: $ \cos {\pi \over 6} = \cos 60^\circ = {\sqrt{3} \over 2} $

\begin{align} \require{cancel} \int_{\pi \over 6}^{\pi} {\cos^2 x \over 1 - \sin x} \phantom{.} dx & = \int_{\pi \over 6}^{\pi} {1 - \sin^2 x \over 1 - \sin x} \phantom{.} dx \phantom{0000000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = \int_{\pi \over 6}^{\pi} { \cancel{(1 - \sin x)}(1 + \sin x) \over \cancel{1 - \sin x} } \phantom{.} dx \phantom{0000000} [a^2 - b^2 = (a + b)(a - b)] \\ & = \int_{\pi \over 6}^{\pi} 1 + \sin x \phantom{.} dx \\ & = \left[ x + {-\cos x \over 1} \right]_{\pi \over 6}^\pi \\ & = \left[ x - \cos x \right]_{\pi \over 6}^\pi \\ & = \pi - \cos \pi - \left( {\pi \over 6} - \cos {\pi \over 6} \right) \\ & = \pi - (-1) - \left( {\pi \over 6} - {\sqrt{3} \over 2} \right) \\ & = \pi + 1 - {\pi \over 6} + {\sqrt{3} \over 2} \\ & = {5\pi \over 6} + 1 + {\sqrt{3} \over 2} \\ & = {5 \over 6}\pi + 1 + {\sqrt{3} \over 2} \end{align}

(d)

\begin{align} {d \over dx} (\cot x) \phantom{0} dx & = {d \over dx} \left( \cos x \over \sin x \right) \\ & = {(\sin x) {d \over dx}(\cos x) - (\cos x){d \over dx}(\sin x) \over (\sin x)^2} \phantom{000000} [\text{Quotient rule}] \\ & = {(\sin x)(-\sin x) - (\cos x)(\cos x) \over \sin^2 x} \\ & = {-\sin^2 x - \cos^2 x \over \sin^2 x} \\ & = {-(\sin^2 x + \cos^2 x) \over \sin^2 x} \\ & = {-(1) \over \sin^2 x} \phantom{0000000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = -{1 \over \sin^2 x} \\ & = -\text{cosec}^2 x \\ \\ \implies \int - \text{cosec}^2 x \phantom{.} dx & = \cot x \\ \\ \therefore \int_{\pi \over 4}^{\pi \over 2} 2 \text{cosec}^2 x \phantom{.} dx & = - 2 \int_{\pi \over 4}^{\pi \over 2} - \text{cosec}^2 x \phantom{.} dx \\ & = -2 [\cot x]_{\pi \over 4}^{\pi \over 2} \\ & = -2 \left[ \cos x \over \sin x \right]_{\pi \over 4}^{\pi \over 2} \\ & = - 2 \left( { \cos {\pi \over 2} \over \sin {\pi \over 2} } - {\cos {\pi \over 4} \over \sin {\pi \over 4}} \right) \\ & = -2(-1) \\ & = 2 \end{align}

(a)

\begin{align} u & = x^2 &&& v & = \sqrt{2x - 1} \\ & &&& & = (2x - 1)^{1 \over 2} \\ {du \over dx} & = 2x &&& {dv \over dx} & = {1 \over 2}(2x - 1)^{-{1 \over 2}}. (2) \phantom{000000} [\text{Chain rule}] \\ & &&& & = (2x - 1)^{-{1 \over 2}} \\ & &&& & = {1 \over \sqrt{2x - 1}} \end{align} \begin{align} {dy \over dx} & = (x^2)\left(1 \over \sqrt{2x - 1}\right) + (\sqrt{2x - 1})(2x) \phantom{000000} [\text{Product rule}] \\ & = {x^2 \over \sqrt{2x - 1}} + 2x \sqrt{2x - 1} \\ & = {x^2 \over \sqrt{2x - 1}} + {2x \sqrt{2x - 1} \over 1} \\ & = {x^2 \over \sqrt{2x - 1}} + {\sqrt{2x - 1} (2x \sqrt{2x - 1}) \over \sqrt{2x - 1}} \\ & = {x^2 \over \sqrt{2x - 1}} + { 2x(2x - 1) \over \sqrt{2x - 1}} \\ & = {x^2 + 2x(2x - 1) \over \sqrt{2x - 1}} \\ & = {x^2 + 4x^2 - 2x \over \sqrt{2x - 1}} \\ & = {5x^2 - 2x \over \sqrt{2x - 1}} \\ & = {x(5x - 2) \over \sqrt{2x - 1}} \phantom{00} \text{ (Shown)} \end{align}

(b)

\begin{align} \text{From (i), } {d \over dx} \left(x^2 \sqrt{2x - 1} \right) & = {x(5x - 2) \over \sqrt{2x - 1}} \\ \\ \implies \int {x(5x - 2) \over \sqrt{2x - 1}} \phantom{.} dx & = x^2 \sqrt{2x - 1} \\ \\ \therefore \int_1^5 {x(5x - 2) \over \sqrt{2x - 1}} \phantom{.} dx & = \left[x^2 \sqrt{2x - 1} \right]_1^5 \\ & = (5)^2 \sqrt{2(5) - 1} - \left[ (1)^2 \sqrt{2(1) - 1} \right] \\ & = (25)(3) - (1)(1) \\ & = 74 \end{align}

(a)

\begin{align} {8x + 13 \over (1 + 2x)(2 + x)^2} & = {A \over 1 + 2x} + {B \over 2 + x} + {C \over (2 + x)^2} \\ {8x + 13 \over (1 + 2x)(2 + x)^2} & = {A(2 + x)^2 \over (1 + 2x)(2 + x)^2} + {B(1 + 2x)(2 + x) \over (1 + 2x)(2 + x)^2} + {C(1 + 2x) \over (1 + 2x)(2 + x)^2} \\ {8x + 13 \over (1 + 2x)(2 + x)^2} & = {A(2 + x)^2 + B(1 + 2x)(2 + x) + C(1 + 2x) \over (1 + 2x)(2 + x)^3} \\ \\ \therefore 8x + 13 & = A(2 + x)^2 + B(1 + 2x)(2 + x) + C(1 + 2x) \\ \\ \text{Let } & x = -2, \\ 8(-2) + 13 & = A(0)^2 + B[1 + 2(-2)](0) + C[1 + 2(-2)] \\ -16 + 13 & = 0 + 0 + C(-3) \\ -3 & = -3C \\ {-3 \over -3} & = C \\ 1 & = C \\ \\ \text{Let } & x = -{1 \over 2}, \\ 8\left(-{1 \over 2}\right) + 13 & = A\left( 2 - {1 \over 2} \right)^2 + B(0)\left(2 - {1 \over 2}\right) + C(0) \\ -4 + 13 & = A\left(9 \over 4\right) + 0 + 0 \\ 9 & = {9 \over 4}A \\ 36 & = 9A \\ {36 \over 9} & = A \\ 4 & = A \\ \\ 8x + 13 & = 4(2 + x)^2 + B(1 + 2x)(2 + x) + (1 + 2x) \\ \\ \text{Let } & x = 0, \\ 8(0) + 13 & = 4(2 + 0)^2 + B[1 + 2(0)](2 + 0) + [1 + 2(0)] \\ 0 + 13 & = 4(4) + B(1)(2) + 1 \\ 13 & = 16 + 2B + 1 \\ -4 & = 2B \\ {-4 \over 2} & = B \\ -2 & = B \\ \\ \therefore {8x + 13 \over (1 + 2x)(2 + x)^2} & = {4 \over 1 + 2x} + {-2 \over 2 + x} + {4 \over (2 + x)^2} \\ & = {4 \over 1 + 2x} - {2 \over 2 + x} + {1 \over (2 + x)^2} \\ \\ \int_1^2 {8x + 13 \over (1 + 2x)(2 + x)^2} \phantom{.} dx & = \int_1^2 {4 \over 1 + 2x} - {2 \over 2 + x} + {1 \over (2 + x)^2} \phantom{.} dx \\ & = \int_1^2 {4 \over 1 + 2x} - {2 \over 2 + x} + (2 + x)^{-2} \phantom{.} dx \\ & = \left[ {4\ln (1 + 2x) \over 2} - {2\ln (2 + x) \over 1} + {(2 + x)^{-1} \over (-1)(1)} \right]_1^2 \\ & = \left[ 2\ln (1 + 2x) - 2\ln (2 + x) -(2 + x)^{-1} \right]_1^2 \\ & = \left[ 2\ln [1 + 2(2)] - 2\ln (2 + 2) - (2 + 2)^{-1} \right] - \left[ 2\ln [1 + 2(1)] - 2\ln (2 + 1) - (2 + 1)^{-1} \right] \\ & = 2 \ln 5 - 2\ln 4 -{1 \over 4} - \left( 2\ln 3 - 2\ln 3 - {1 \over 3} \right) \\ & = 2\ln 5 - 2\ln 4 - {1 \over 4} - \left( - {1 \over 3} \right) \\ & = 2\ln 5 - 2\ln 4 - {1 \over 4} + {1 \over 3} \\ & = 2\ln 5 - 2\ln 4 + {1 \over 12} \\ & = 2(\ln 5 - \ln 4) + {1 \over 12} \\ & = 2\ln {5 \over 4} + {1 \over 12} \phantom{000000000000} [\text{Quotient law (logarithms)}] \end{align}

(b)

\begin{align} {1 \over x^2 + 3x + 2} & = {1 \over (x + 2)(x +1)} \\ {1 \over x^2 + 3x + 2} & = {A \over x + 2} + {B \over x + 1} \\ {1 \over x^2 + 3x + 2} & = {A(x + 1) \over (x + 2)(x + 1)} + {B(x + 2) \over (x + 2)(x + 1)} \\ {1 \over x^2 + 3x + 2} & = {A(x + 1) + B(x + 2) \over (x + 2)(x + 1)} \\ \\ \therefore 1 & = A(x + 1) + B(x + 2) \\ \\ \text{Let } & x = -1, \\ 1 & = A(0) + B(-1 + 2) \\ 1 & = 0 + B(1) \\ 1 & = B \\ \\ \text{Let } & x = - 2, \\ 1 & = A(-2 + 1) + B(0) \\ 1 & = A(-1) + 0 \\ 1 & = -A \\ -1 & = A \\ \\ \therefore {1 \over (x + 2)(x + 1)} & = {A \over x + 2} + {B \over x + 1} \\ & = {-1 \over x + 2} + {1 \over x + 1} \\ & = -{1 \over x + 2} + {1 \over x + 1} \\ \\ \int {1 \over (x + 2)(x + 1)} \phantom{.} dx & = \int -{1 \over x + 2} + {1 \over x + 2} \phantom{.} dx \\ & = - {\ln (x + 2) \over 1} + {\ln (x + 1) \over 1} \\ & = -\ln (x + 2) + \ln (x + 1) \\ & = \ln (x + 1) - \ln (x + 2) \\ & = \ln \left( {x + 1 \over x + 2} \right) + c \phantom{00000000000} [\text{Quotient law (logarithms)}] \end{align}

Differentiation part

\begin{align} {d \over dx} (\sin^3 2x) & = {d \over dx} (\sin 2x)^3 \\ & = (3)(\sin 2x)^2 . (2)(\cos 2x) \phantom{000000} [\text{Chain rule}] \\ & = 6 \sin^2 2x \cos 2x \end{align}

(i)

\begin{align} {d \over dx} (\sin^3 2x) & = 6 \sin^2 2x \cos 2x \\ \\ \implies \int 6 \sin^2 2x \cos 2x \phantom{.} dx & = \sin^3 2x \\ \\ \therefore \int_0^{\pi \over 4} \sin^2 2x \cos 2x \phantom{.} dx & = {1 \over 6} \int_0^{\pi \over 4} 6 \sin^2 2x \cos 2x \phantom{.} dx \\ & = {1 \over 6} \left[ \sin^3 2x \right]_0^{\pi \over 4} \\ & = {1 \over 6} \left\{ \sin^3 \left[2 \left({\pi \over 4}\right) \right] - \sin^3 [2(0)] \right\} \\ & = {1 \over 6} \left( \sin^3 {\pi \over 2} - \sin^3 0 \right) \\ & = {1 \over 6} \left[ (1)^3 - {1 \over 6} (0)^3 \right] \\ & = {1 \over 6} (1) \\ & = {1 \over 6} \end{align}

(ii)

\begin{align} \therefore \int_0^{\pi \over 4} \cos^3 2x \phantom{.} dx & = \int_0^{\pi \over 4} \cos 2x (\cos^2 2x) \phantom{.} dx \\ & = \int_0^{\pi \over 4} \cos 2x (1 - \sin^2 2x) \phantom{.} dx \phantom{000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = \int_0^{\pi \over 4} \cos 2x - \sin^2 2x \cos 2x \phantom{.} dx \\ & = \int_0^{\pi \over 4} \cos 2x \phantom{.} dx - \int_0^{\pi \over 4} \sin^2 2x \cos 2x \phantom{.} dx \\ & = \int_0^{\pi \over 4} \cos 2x \phantom{.} dx - {1 \over 6} \phantom{000000000000000000} [\text{Use answer from (i)}] \\ & = \left[ \sin 2x \over 2\right]_0^{\pi \over 4} - {1 \over 6} \\ & = { \sin \left[2 \left(\pi \over 4\right) \right] \over 2} - { \sin [2(0)] \over 2} - {1 \over 6} \\ & = { 1 \over 2} - {0 \over 2} - {1 \over 6} \\ & = {1 \over 3} \end{align}

(a)

\begin{align} {d \over dx} [\ln (\ln x)] & = { {1 \over x} \over \ln x } \\ & = {1 \over x} \div \ln x \\ & = {1 \over x} \times {1 \over \ln x} \\ & = {1 \over x\ln x} \\ \\ \\ {d \over dx} [\ln (\ln x)] & = {1 \over x \ln x} \\ \\ \implies \int {1 \over x \ln x} \phantom{.} dx & = \ln (\ln x) \\ \\ \therefore \int_2^4 {1 \over x\ln x} \phantom{.} dx & = [\ln (\ln x)]_2^4 \\ & = \ln (\ln 4) - \ln (\ln 2) \\ & = \ln \left( \ln 4 \over \ln 2 \right) \phantom{0000000000} [\text{Quotient law (logarithms)}] \\ & = \ln 2 \end{align}

(b)

\begin{align} u & = x^n &&& v & = \ln x \\ {du \over dx} & = n x^{n - 1} &&& {dv \over dx} & = {1 \over x} \end{align} \begin{align} {d \over dx} (x^n \ln x) & = (x^n)\left(1 \over x\right) + (\ln x)(n x^{n - 1}) \phantom{000000} [\text{Product rule}] \\ & = {x^n \over x} + n x^{n - 1} \ln x \\ & = x^{n - 1} + n x^{n - 1} \ln x \\ \\ \implies \int x^{n - 1} + n x^{n - 1} \ln x \phantom{.} dx & = x^n \ln x \\ \int x^{n - 1} \phantom{.} dx + \int n x^{n - 1} \ln x \phantom{.} dx & = x^n \ln x \\ \int n x^{n - 1} \ln x \phantom{.} dx & = x^n \ln x - \int x^{n - 1} \phantom{.} dx \\ & = x^n \ln x - {x^n \over n} \\ \\ \\ \therefore \text{General case: } \int n x^{n - 1} \ln x \phantom{.} dx & = x^n \ln x - {x^n \over n} \\ \\ \text{When } & n = 1, \\ \int (1) x^{1 - 1} \ln x \phantom{.} dx & = x^1 \ln x - {x^1 \over 1} \\ \int (1) x^0 \ln x \phantom{.} dx & = x \ln x - x \\ \int (1) (1) \ln x \phantom{.} dx & = x \ln x - x \\ \int \ln x \phantom{.} dx & = x \ln x - x + c \\ \\ \text{When } & n = 2, \\ \int (2) x^{2 - 1} \ln x \phantom{.} dx & = x^2 \ln x - {x^2 \over 2} \\ 2 \int x \ln x \phantom{.} dx & = x^2 \ln x - {x^2 \over 2} \\ \int x \ln x \phantom{.} dx & = {1 \over 2} \left( x^2 \ln x - {x^2 \over 2} \right) + c \\ \\ \text{When } & n = 3, \\ \int (3) x^{3 - 1} \ln x \phantom{.} dx & = x^3 \ln x - {x^3 \over 3} \\ 3 \int x^2 \ln x \phantom{.} dx & = x^3 \ln x - {x^3 \over 3} \\ \int x^2 \ln x \phantom{.} dx & = {1 \over 3} \left( x^3 \ln x - {x^3 \over 3} \right) + c \\ \\ \text{When } & n = m + 1, \\ \int (m + 1)x^{m + 1 - 1} \ln x \phantom{.} dx & = x^{m + 1} \ln x - {x^{m + 1} \over m + 1} \\ (m + 1) \int x^m \ln x \phantom{.} dx & = x^{m + 1} \ln x - {x^{m + 1} \over m + 1} \\ \int x^m \ln x \phantom{.} dx & = {1 \over m + 1} \left( x^{m + 1} \ln x - {x^{m + 1} \over m + 1} \right) + c \end{align}

(a)

\begin{align} \int {1 \over N} \phantom{.} dN & = kt \\ \ln N + c & = kt \\ \\ \text{When } t = 0 & \text{ and } N = A, \phantom{000000} [\text{Initial bacteria is } A] \\ \ln A + c & = 0 \\ c & = - \ln A \\ \\ \ln N - \ln A & = kt \\ \ln {N \over A} & = kt \phantom{0000000000} [\text{Quotient law (logarithms)}] \\ \ln {N \over A} & = kt \ln e \\ \ln {N \over A} & = \ln e^{kt} \phantom{000000(.} [\text{Power law (logarithms)}] \\ \\ \implies {N \over A} & = e^{kt} \\ N & = Ae^{kt} \phantom{00} \text{ (Shown)} \end{align}

(b) Note: I think this part is not required for O levels. It's from A level H2 maths - topic is Differential Equations

(i) Solve by variable separable method

\begin{align} {dr \over dt} & = - {r \over 10} \\ {1 \over r} \cdot {dr \over dt} & = -{1 \over 10} \\ \int {1 \over r} \cdot {dr \over dt} \phantom{.} dt & = \int -{1 \over 10} \phantom{.} dt \\ \int {1 \over r} \phantom{.} dr & = -{1 \over 10}t + c \\ \ln r & = -{1 \over 10}t + c \\ \\ \text{When } t = 0 & \text{ and } r = 8, \phantom{000000} [\text{Initial radius is 8 cm}] \\ \ln 8 & = -{0 \over 10} + c \\ \ln 8 & = 0 + c \\ \ln 8 & = c \\ \\ \ln r & = -{t \over 10} + \ln 8 \\ \ln r - \ln 8 & = -{t \over 10} \\ \ln {r \over 8} & = -{t \over 10} \\ \log_e {r \over 8} & = -{t \over 10} \\ {r \over 8} & = e^{-{t \over 10}} \\ r & = 8e^{-{t \over 10}} \end{align}

(ii)

\begin{align} r & = 8e^{-{t \over 10}} \\ \\ \text{When } & r = 4, \\ 4 & = 8e^{-{t \over 10}} \\ {4 \over 8} & = e^{-{ t \over 10}} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln {4 \over 8} & = \ln e^{-{t \over 10}} \\ \ln {1 \over 2} & = -{t \over 10} \ln e \\ \ln {1 \over 2} & = -{t \over 10} (1) \\ \ln {1 \over 2} & = -{t \over 10} \\ 10\ln {1 \over 2} & = -t \\ -10 \ln {1 \over 2} & = t \\ \\ t & = -10 \ln {1 \over 2} \\ & = 6.9314 \\ & \approx 6.93 \text{ min} \end{align}