\begin{align*} \int_0^1 e^{2x} (e^x + e^{-x}) \phantom{.} dx & = \int_0^1 e^{3x} + e^x \phantom{.} dx \phantom{000000} [a^m \times a^n = a^{m + n} ] \\ & = \left[ {e^{3x} \over 3} + {e^x \over 1} \right]_0^1 \\ & = { e^{3(1)} \over 3} + e^1 - \left[ {e^{3(0)} \over 3} + e^0 \right] \\ & = { e^3 \over 3 } + e - {1 \over 3} - 1 \\ & = { e^3 \over 3} + {3e \over 3} - {4 \over 3} \\ & = { e^3 + 3e - 4 \over 3} \\ \\ \therefore a & = 3, b = -4 \end{align*}

\begin{align*} {x \over (x + 1)(x + 2)^2} & = {A \over x + 1} + {B \over x + 2} + {C \over (x + 2)^2} \\ & = {A(x + 2)^2 \over (x + 1)(x + 2)^2} + {B(x + 1)(x + 2) \over (x + 1)(x + 2)^2} + {C(x + 1) \over (x + 1)(x + 2)^2} \\ & = {A(x + 2)^2 + B(x + 1)(x + 2) + C(x + 1) \over (x + 1)(x + 2)^2 } \\ \\ \therefore x & = A(x + 2)^2 + B(x + 1)(x + 2) + C(x + 1) \\ \\ \text{Let } & x = -2, \\ -2 & = A(0)^2 + B(-1)(0) + C(-1) \\ -2 & = -C \\ 2 & = C \\ \\ x & = A(x + 2)^2 + B(x + 1)(x + 2) + 2(x + 1) \\ \\ \text{Let } & x = -1, \\ -1 & = A(1)^2 + B(0)(1) + 2(0) \\ -1 & = A + 0 + 0 \\ -1 & = A \\ \\ x & = -(x + 2)^2 + B(x + 1)(x + 2) + 2(x + 1) \\ \\ \text{Let } & x = 0, \\ 0 & = -(2)^2 + B(1)(2) + 2(1) \\ 0 & = -4 + 2B + 2 \\ 2 & = 2B \\ 1 & = B \\ \\ \therefore {x \over (x + 1)(x + 2)^2} & = {-1 \over x + 1} + {1 \over x + 2} + {2 \over (x + 2)^2} \\ & = - {1 \over x + 1} + {1 \over x + 2} + {2 \over (x + 2)^2} \end{align*}

\begin{align*} \int {x \over (x + 1)(x + 2)^2} \phantom{.} dx & = \int - {1 \over x + 1} + {1 \over x + 2} + {2 \over (x + 2)^2} \phantom{.} dx \\ & = \int - {1 \over x + 1} + {1 \over x + 2} + 2(x + 2)^{-2} \phantom{.} dx \\ & = - { \ln (x + 1) \over 1} + {\ln (x + 2) \over 1} + {2 (x + 2)^{-1} \over (-1)(1)} \\ & = - \ln (x + 1) + \ln (x + 2) - 2(x + 2)^{-1} \\ & = \ln (x + 2) - \ln (x + 1) - {2 \over x + 2} \\ & = \ln {x +2 \over x + 1} - {2 \over x + 2} \phantom{000000} [\text{Quotient law (logarithms)}] \\ \\ \therefore \int_0^1 {x \over (x + 1)(x + 2)^2} \phantom{.} dx & = \left[ \ln {x +2 \over x + 1} - {2 \over x + 2} \right]_0^1 \\ & = \ln {3 \over 2} - {2 \over 3} - \left( \ln 2 - 1 \right) \\ & = \ln {3 \over 2} - \ln 2 - {2 \over 3} + 1 \\ & = \ln { {3 \over 2} \over 2 } + {1 \over 3} \\ & = \ln {3 \over 4} + {1 \over 3} \phantom{0} \text{ (Shown)} \end{align*}

\begin{align*} 4 \sin^2 x + 6 \cos^2 x & = 2 (2 \sin^2 x) + 3(2 \cos^2 x) \\ & = 2 (1 - \cos 2x) + 3(2 \cos^2 x) \phantom{000000} [ \cos 2A = 1 - 2 \sin^2 A \implies 2 \sin^2 A = 1 - \cos 2A] \\ & = 2 - 2 \cos 2x + 3 (\cos 2x + 1) \phantom{00000} [\cos 2A = 2 \cos^2 A - 1 \implies 2 \cos^2 A = \cos 2A + 1] \\ & = 2 - 2 \cos 2x + 3 \cos 2x + 3 \\ & = \cos 2x + 5 \phantom{0} \text{ (Shown)} \end{align*}

\begin{align*} \int_{\pi \over 6}^{\pi \over 2} 4 \sin^2 x + 6 \cos^2 x \phantom{.} dx & = \int_{\pi \over 6}^{\pi \over 2} \cos 2x + 5 \phantom{.} dx \\ & = \left[ {\sin 2x \over 2} + 5x \right]_{\pi \over 6}^{\pi \over 2} \\ & = \left[ {1 \over 2} \sin 2x + 5x \right]_{\pi \over 6}^{\pi \over 2} \\ & = {1 \over 2} \sin \pi + 5 \left(\pi \over 2\right) - \left[ {1 \over 2} \sin {\pi \over 3} + 5 \left(\pi \over 6\right) \right] \\ & = {1 \over 2} (0) + {5 \over 2} \pi - {1 \over 2} \left(\sqrt{3} \over 2\right) - {5 \over 6} \pi \\ & = {5 \over 2} \pi - {5 \over 6} \pi - {\sqrt{3} \over 4} \\ & = {5 \over 3} \pi - {\sqrt{3} \over 4} \phantom{0} \text{ (Shown)} \end{align*}

\begin{align*} 2 \tan^2 {1 \over 2}x - 3 & = 2 \left( \sec^2 {1 \over 2}x - 1 \right) - 3 \phantom{000000} [\tan^2 A + 1 = \sec^2 A \implies \tan^2 A = \sec^2 A - 1] \\ & = 2 \sec^2 {1 \over 2}x - 2 - 3 \\ & = 2 \sec^2 {1 \over 2}x - 5 \\ \\ \therefore \int 2 \sec^2 {1 \over 2}x - 5 \phantom{.} dx & = {2 \tan {1 \over 2}x \over {1 \over 2} } - 5x \\ & = 4 \tan {1 \over 2}x - 5x + c \end{align*}

\begin{align*} f'(x) & = \sin 6x + \cos 3x \\ \\ f''(x) & = 6 \cos 6x + (3)(-\sin 3x) \phantom{000000} [\text{Differentiate } f'(x)] \\ & = 6 \cos 6x - 3 \sin 3x \\ \\ f(x) & = \int f'(x) \phantom{.} dx \\ & = \int \sin 6x + \cos 3x \phantom{.} dx \\ & = {-\cos 6x \over 6} + {\sin 3x \over 3} \\ & = - {1 \over 6} \cos 6x + {1 \over 3} \sin 3x + c \\ \\ \text{Since } & f \left({\pi \over 6}\right) = {5 \over 6}, \\ {5 \over 6} & = -{1 \over 6} \cos \pi + {1 \over 3} \sin {\pi \over 2} + c \\ {5 \over 6} & = {1 \over 6} + {1 \over 3} + c \\ {1 \over 3} & = c \\ \\ f(x) & = - {1 \over 6} \cos 6x + {1 \over 3} \sin 3x + {1 \over 3} \\ \\ \\ f''(x) + 9 f(x) & = 6 \cos 6x - 3 \sin 3x + 9 \left( - {1 \over 6} \cos 6x + {1 \over 3} \sin 3x + {1 \over 3} \right) \\ & = 6 \cos 6x - 3 \sin 3x - {3 \over 2} \cos 6x + 3 \sin 3x + 3 \\ & = {9 \over 2} \cos 6x + 3 \phantom{0} \text{ (Shown)} \end{align*}