(a)

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

(b)

\begin{align} \int 2e^{3x} \phantom{.} dx & = 2 \left( {e^{3x} \over 3} \right) \\ & = {2 \over 3}e^{3x} + c \end{align}

(c)

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

(d)

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

(e)

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

(f)

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

(a)

\begin{align} \int_0^2 e^x \phantom{.} dx & = \left[e^x \over 1\right]_0^2 \\ & = [e^x]_0^2 \\ & = e^2 - e^0 \\ & = e^2 - 1 \end{align}

(b)

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

(c)

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

(d)

\begin{align} \int_1^2 e^{1 - x} \phantom{.} dx & = \left[ e^{1 - x} \over - 1 \right]_1^2 \\ & = \left[ - e^{1 - x} \right]_1^2 \\ & = - e^{1 - 2} - \left( -e^{1 - 1} \right) \\ & = - e^{-1} - ( - e^0 ) \\ & = -e^{-1} - (- 1) \\ & = -{1 \over e} + 1 \\ & = 1 - {1 \over e} \end{align}

(e)

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

(f)

\begin{align} \int_0^{\ln 2} 5e^{3x} \phantom{.} dx & = \left[ 5e^{3x} \over 3 \right]_0^{\ln 2} \\ & = \left[ {5 \over 3} e^{3x} \right]_0^{\ln 2} \\ & = {5 \over 3} e^{3(\ln 2)} - {5 \over 3} e^{3(0)} \\ & = {5 \over 3} e^{3\ln 2} - {5 \over 3} (1) \\ & = {5 \over 3} (8) - {5 \over 3} \\ & = 11{2 \over 3} \end{align}

\begin{align} y & = \int {dy \over dx} \phantom{.} dx \\ y & = \int e^{2x} \phantom{.} dx \\ y & = {e^{2x} \over 2} \\ y & = {1 \over 2}e^{2x} + c \\ \\ \text{When } & x = 1 \text{ and } y = 2, \\ 2 & = {1 \over 2}e^{2(1)} + c \\ 2 & = {1 \over 2}e^2 + c \\ 2 - {1 \over 2}e^2 & = c \\ \\ \therefore y & = {1 \over 2}e^{2x} + 2 - {1 \over 2}e^2 \end{align}

(a)

\begin{align} \int {2 \over x} \phantom{.} dx & = 2\int {1 \over x} \phantom{.} dx \\ & = 2\left(\ln x \over 1\right) \\ & = 2\ln x + c \end{align}

(b)

\begin{align} \int {1 \over x + 1} \phantom{.} dx & = {\ln (x + 1) \over 1} \\ & = \ln (x + 1) + c \end{align}

(c)

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

(d)

\begin{align} \int {3 \over 2x + 1} \phantom{.} dx & = 3 \int {1 \over 2x + 1} \phantom{.} dx \\ & = 3\left[ \ln (2x + 1) \over 2 \right] \\ & = {3 \over 2} \ln (2x + 1) + c \end{align}

(e)

\begin{align} \int {2 \over 1 - 2x} \phantom{.} dx & = 2 \int {1 \over 1 - 2x} \phantom{.} dx \\ & = 2 \left[ \ln (1 - 2x) \over - 2 \right] \\ & = -\ln (1 - 2x) + c \end{align}

(f)

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

(a)

\begin{align} \int_1^4 {4 \over x} \phantom{.} dx & = 4 \int_1^4 {1 \over x} \phantom{.} dx \\ & = 4 \left[ \ln x \over 1 \right]_1^4 \\ & = 4 \left[ \ln x \right]_1^4 \\ & = 4 \left[ \ln 4 - \ln 1 \right] \\ & = 4 \left[ \ln 4 - 0 \right] \\ & = 4\ln 4 \\ & = 4\ln 2^2 \\ & = 4(2 \ln 2) \phantom{0000000000} [\text{Power law (logarithms)}] \\ & = 8 \ln 2 \end{align}

(b)

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

(c)

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

(d)

\begin{align} \int_0^1 {3 \over 2 + 3x} \phantom{.} dx & = \left[ 3\ln (2 + 3x) \over 3 \right]_0^1 \\ & = \left[\ln (2 + 3x) \right]_0^1 \\ & = \ln [2 + 3(1)] - \ln [2 + 3(0)] \\ & = \ln 5 - \ln 2 \\ & = \ln {5 \over 2} \phantom{0000000000} [\text{Quotient law (logarithms)}] \end{align}

(a)

\begin{align} y & = \int {dy \over dx} \phantom{.} dx \\ y & = \int {1 \over 2x + 1} \phantom{.} dx \\ y & = {\ln (2x + 1) \over 2} \\ y & = {1 \over 2}\ln (2x + 1) + c \\ \\ \text{When } & x = 0 \text{ and } y = 0.5, \\ 0.5 & = {1 \over 2}\ln [2(0) + 1] + c \\ 0.5 & = {1 \over 2}\ln 1 + c \\ 0.5 & = 0 + c \\ 0.5 & = c \\ {1 \over 2} & = c \\ \\ \therefore y & = {1 \over 2}\ln (2x + 1) + {1 \over 2} \end{align}

(b)

\begin{align} y & = \int {dy \over dx} \phantom{.} dx \\ y & = \int {1 \over 3 - 2x} \phantom{.} dx \\ y & = {\ln (3 - 2x) \over -2} \\ y & = -{1 \over 2} \ln (3 - 2x) + c \\ \\ \text{When } & x = 1 \text{ and } y = 2, \\ 2 & = -{1 \over 2} \ln [3 - 2(1)] + c \\ 2 & = -{1 \over 2}\ln 1 + c \\ 2 & = 0 + c \\ 2 & = c \\ \\ \therefore y & = -{1 \over 2}\ln (3 - 2x) + 2 \\ y & = 2 - {1 \over 2}\ln (3 - 2x) \end{align}

(a)

\begin{align} & \phantom{...} \int_0^1 \left( e^{{2 \over 3}x} + e^{-{2 \over 3}x} \right )^2 \phantom{.} dx \\ & = \int_0^1 \left(e^{{2 \over 3}x}\right)^2 + 2\left(e^{{2 \over 3}x}\right)\left(e^{-{2 \over 3}x}\right) + \left(e^{-{2 \over 3}x}\right)^2 \phantom{.} dx \phantom{0000000000} [(a + b)^2 = a^2 + 2ab + b^2] \\ & = \int_0^1 e^{{4 \over 3}x} + 2 + e^{-{4 \over 3}x} \phantom{.} dx \\ & = \left[ {e^{{4 \over 3}x} \over {4 \over 3}} + 2x + {e^{-{4 \over 3}x} \over -{4 \over 3}} \right]_0^1 \\ & = \left[ {3 \over 4}e^{{4 \over 3}x} + 2x - {3 \over 4}e^{-{4 \over 3}x} \right]_0^1 \\ & = {3 \over 4}e^{{4 \over 3}(1)} + 2(1) - {3 \over 4}e^{-{4 \over 3}(1)} - \left[ {3 \over 4}e^{{4 \over 3}(0)} + 2(0) - {3 \over 4}e^{-{4 \over 3}(0)} \right] \\ & = {3 \over 4}e^{4 \over 3} + 2 - {3 \over 4}e^{-{4 \over 3}} - \left[ {3 \over 4}e^0 + 0 - {3 \over 4}e^0 \right] \\ & = {3 \over 4}e^{4 \over 3} + 2 - {3 \over 4}e^{-{4 \over 3}} - \left[ {3 \over 4}(1) - {3 \over 4}(1) \right] \\ & = {3 \over 4}e^{4 \over 3} + 2 - {3 \over 4}e^{-{4 \over 3}} - 0 \\ & = {3 \over 4}e^{4 \over 3} - {3 \over 4}e^{-{4 \over 3}} + 2 \\ & = {3 \over 4} \left( e^{4 \over 3} - e^{-{4 \over 3}} \right) + 2 \end{align}

(b)

\begin{align} \int_0^1 {e^x + e^{2x} \over e^{-x}} \phantom{.} dx & = \int_0^1 {e^x \over e^{-x}} + {e^{2x} \over e^{-x}} \phantom{.} dx \\ & = \int_0^1 e^{x - (- x)} + e^{2x - (-x)} \phantom{0} dx \\ & = \int_0^1 e^{2x} + e^{3x} \phantom{.} dx \\ & = \left[ {e^{2x} \over 2} + {e^{3x} \over 3} \right]_0^1 \\ & = \left[ {1 \over 2}e^{2x} + {1 \over 3}e^{3x} \right]_0^1 \\ & = {1 \over 2}e^{2(1)} + {1 \over 3}e^{3(1)} - \left[ {1 \over 2}e^{2(0)} + {1 \over 3}e^{3(0)} \right] \\ & = {1 \over 2}e^2 + {1 \over 3}e^3 - \left( {1 \over 2}e^0 + {1 \over 3}e^0 \right) \\ & = {1 \over 2}e^2 + {1 \over 3}e^3 - \left[ {1 \over 2}(1) + {1 \over 3}(1) \right] \\ & = {1 \over 2}e^2 + {1 \over 3}e^3 - {5 \over 6} \end{align}

(i)

\begin{align} {2x - 1 \over (x + 1)(x + 2)} & = {A \over x + 1} + {B \over x + 2} \\ {2x - 1 \over (x + 1)(x + 2)} & = {A(x + 2) \over (x + 1)(x + 2)} + {B(x + 1) \over (x + 1)(x + 2)} \\ {2x - 1 \over (x + 1)(x + 2)} & = {A(x + 2) + B(x + 1) \over (x + 1)(x + 2)} \\ \\ \therefore 2x - 1 & = A(x + 2) + B(x + 1) \\ \\ \text{Let } & x = - 2, \\ 2(-2) - 1 & = A(0) + B(-2 + 1) \\ -4 - 1 & = 0 + B(-1) \\ -5 & = - B \\ 5 & = B \\ \\ \text{Let } & x = -1, \\ 2(-1) - 1 & = A[(-1) + 2) + B(0) \\ -2 - 1 & = A(1) + 0 \\ - 3 & = A \\ \\ \therefore {2x - 1 \over (x + 1)(x + 2)} & = {-3 \over x + 1} + {5 \over x + 2} \end{align}

(ii)

\begin{align} \int_1^2 {2x - 1 \over (x + 1)(x + 2)} \phantom{.} dx & = \int_1^2 {-3 \over x + 1} + {5 \over x + 2} \phantom{.} dx \\ & = \left[ {-3\ln (x + 1) \over 1} + {5 \ln (x + 2) \over 1} \right]_1^2 \\ & = \left[ -3\ln (x + 1) + 5\ln (x + 2) \right]_1^2 \\ & = -3\ln (2 + 1) + 5\ln (2 + 2) - \left[ -3\ln (1 + 1) + 5\ln (1 + 2) \right] \\ & = -3\ln 3 + 5\ln 4 - \left( -3\ln 2 + 5\ln 3 \right) \\ & = -3\ln 3 + 5\ln 4 + 3\ln 2 - 5\ln 3 \\ & = 5\ln 4 + 3\ln 2 - 8\ln 3 \\ & = 5\ln 2^2 + 3\ln 2 - 8\ln 3 \\ & = 2(5 \ln 2) + 3\ln 2 - 8\ln 3 \phantom{0000000000} [\text{Power law (logarithms)}] \\ & = 10 \ln 2 + 3\ln 2 - 8 \ln 3 \\ & = 13 \ln 2 - 8\ln 3 \end{align}

(i)

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

(ii)

\begin{align} \int {1 \over (1 - x)(3 - 2x) } \phantom{.} dx & = \int {1 \over 1 - x} - {2 \over 3 - 2x} \phantom{.} dx \\ & = {\ln (1 - x) \over -1} - {2 \ln (3 - 2x) \over - 2} + c \\ & = -\ln (1 - x) + \ln (3 - 2x) + c \\ & = \ln (3 - 2x) - \ln (1 - x) + c \\ & = \ln \left(3 - 2x \over 1 - x\right) + c \phantom{00000000} [\text{Quotient law (logarithms)}] \end{align}

To integrate ${dy \over dx}$, we need to express the fraction in partial fractions.

\begin{align} {1 \over (x - 1)(x - 2)} & = {A \over x - 1} + {B \over x - 2} \\ {1 \over (x - 1)(x - 2)} & = {A(x - 2) \over (x - 1)(x - 2)} + {B(x - 1) \over (x - 1)(x - 2)} \\ {1 \over (x - 1)(x - 2)} & = {A(x - 2) + B(x - 1) \over (x - 1)(x - 2)} \\ \\ \therefore 1 & = A(x - 2) + B(x - 1) \\ \\ \text{Let } & x = 2, \\ 1 & = A(0) + B(2 - 1) \\ 1 & = 0 + B(1) \\ 1 & = B \\ \\ \text{Let } & x = 1, \\ 1 & = A(1 - 2) + B(0) \\ 1 & = A(-1) + 0 \\ 1 & = -A \\ -1 & = A \\ \\ \therefore {dy \over dx} & = {1 \over (x - 1)(x - 2) } \\ & = {-1 \over x - 1} + {1 \over x - 2} \\ \\ y & = \int {dy \over dx} \phantom{.} dx \\ y & = \int {-1 \over x - 1} + {1 \over x - 2} \phantom{.} dx \\ y & = {-\ln (x - 1) \over 1} + {\ln (x - 2) \over 1} + c \\ y & = -\ln (x - 1) + \ln (x - 2) + c \\ \\ \text{When } & x = 3 \text{ and } y = 2, \\ 2 & = -\ln (3 - 1) + \ln (3 - 2) + c \\ 2 & = -\ln 2 + \ln 1 + c \\ 2 & = -\ln 2 + 0 + c \\ 2 + \ln 2 & = c \\ \\ \therefore y & = -\ln (x - 1) + \ln (x - 2) + 2 + \ln 2 \\ & = \ln (x - 2) + \ln 2 - \ln (x - 1) + 2 \\ & = \ln [2(x - 2)] - \ln (x - 1) + 2 \phantom{000000000} [\text{Product law (logarithms)}] \\ & = \ln (2x - 4) - \ln (x - 1) + 2 \\ & = \ln \left(2x - 4 \over x - 1\right) + 2 \phantom{0000000000000000.} [\text{Quotient law (logarithms)}] \end{align}

(i)

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

(ii)

\begin{align} \int_3^5 {1 \over (x - 2)(x - 1)^2} \phantom{.} dx & = \int_3^5 {1 \over x - 2} - {1 \over x - 1} - {1 \over (x - 1)^2} \phantom{.} dx \\ & = \int_3^5 {1 \over x - 2} - {1 \over x - 1} - (x - 1)^{-2} \phantom{.} dx \\ & = \left[ {\ln (x - 2) \over 1} - {\ln (x - 1) \over 1} - {(x -1)^{-1} \over (-1)(1)} \right]_3^5 \\ & = \left[ \ln (x - 2) - \ln (x - 1) + (x - 1)^{-1} \right]_3^5 \\ & = \left[ \ln (x - 2) - \ln (x - 1) + {1 \over x - 1} \right]_3^5 \\ & = \ln (5 - 2) - \ln (5 - 1) + {1 \over 5 - 1} - \left[ \ln (3 - 2) - \ln (3 - 1) + {1 \over 3 -1} \right] \\ & = \ln 3 - \ln 4 + {1 \over 4} - \left[ \ln 1 - \ln 2 + {1 \over 2} \right] \\ & = \ln 3 - \ln 4 + {1 \over 4} - \left[ 0 - \ln 2 + {1 \over 2} \right] \\ & = \ln 3 - \ln 4 + {1 \over 4} + \ln 2 - {1 \over 2} \\ & = \ln 3 - \ln 4 + \ln 2 - {1 \over 4} \\ & = \ln {3 \over 4} + \ln 2 - {1 \over 4} \phantom{000000000000} [\text{Quotient law (logarithms)}] \\ & = \ln \left( {3 \over 4} \times 2 \right) - {1 \over 4} \phantom{00000000000.} [\text{Product law (logarithms)}] \\ & = \ln {3 \over 2} - {1 \over 4} \end{align}

(i)

\begin{align} {1 \over (x + 3)(x - 2)^2} & = {A \over x + 3} + {B \over x - 2} + {C \over (x - 2)^2} \\ {1 \over (x + 3)(x - 2)^2} & = {A(x -2)^2 \over (x + 3)(x - 2)^2} + {B(x + 3)(x -2) \over (x + 3)(x - 2)^2} + {C(x + 3) \over (x + 3)(x - 2)^2} \\ {1 \over (x + 3)(x - 2)^2} & = {A(x - 2)^2 + B(x + 3)(x - 2) + C(x + 3) \over (x + 3)(x - 2)^2} \\ \\ \therefore 1 & = A(x - 2)^2 + B(x + 3)(x - 2) + C(x + 3) \\ \\ \text{Let } & x = 2, \\ 1 & = A(0)^2 + B(2 + 3)(0) + C(2 + 3) \\ 1 & = 0 + 0 + C(5) \\ 1 & = 5C \\ {1 \over 5} & = C \\ \\ \text{Let } & x = -3, \\ 1 & = A(-3 -2)^2 + B(0)(-3 -2) + C(0) \\ 1 & = A(-5)^2 + 0 + 0 \\ 1 & = A(25) \\ 1 & = 25A \\ {1 \over 25} & = A \\ \\ \therefore 1 & = A(x - 2)^2 + B(x - 3)(x - 2) + C(x + 3) \\ 1 & = {1 \over 25}(x - 2)^2 + B(x - 3)(x - 2) + {1 \over 5}(x + 3) \\ \\ \text{Let } & x = 0, \\ 1 & = {1 \over 25}(0 - 2)^2 + B(0 + 3)(0 - 2) + {1 \over 5}(0 + 3) \\ 1 & = {1 \over 25}(4) + B(3)(-2) + {1 \over 5}(3) \\ 1 & = {4 \over 25} + B(-6) + {3 \over 5} \\ 1 - {4 \over 25} - {3 \over 5} & = -6B \\ {6 \over 25} & = -6B \\ -{6 \over 25} & = 6B \\ -{1 \over 25} & = B \\ \\ \therefore {1 \over (x + 3)(x - 2)^2} & = {A \over x + 3} + {B \over x - 2} + {C \over (x - 2)^2} \\ & = {{1 \over 25} \over x + 3} + {-{1 \over 25} \over x - 2} + {{1 \over 5} \over (x - 2)^2} \\ & = {1 \over 25(x + 3)} - {1 \over 25(x - 2)} + {1 \over 5(x - 2)^2} \\ & = {1 \over 25(x + 3)} - {1 \over 25(x - 2)} + {5 \over 25(x - 2)^2} \\ & = {1 \over 25} \left[ {1 \over x + 3} - {1 \over x - 2} + {5 \over (x - 2)^2} \right] \end{align}

(ii)

\begin{align} \int_3^5 {1 \over (x + 3)(x - 2)^2} \phantom{.} dx & = \int_3^5 {1 \over 25} \left[ {1 \over x + 3} - {1 \over x - 2} + {5 \over (x - 2)^2} \right] \phantom{.} dx \\ & = {1 \over 25} \int_3^5 {1 \over x + 3} - {1 \over x - 2} + 5(x - 2)^{-2} \phantom{.} dx \\ & = {1 \over 25} \left[ {\ln (x + 3) \over 1} - {\ln (x - 2) \over 1} + {5(x - 2)^{-1} \over (-1)(1)} \right]_3^5 \\ & = {1 \over 25} \left[ \ln (x + 3) - \ln (x - 2) - 5(x - 2)^{-1} \right]_3^5 \\ & = {1 \over 25} \left[ \ln (x + 3) - \ln (x - 2) - {5 \over x - 2} \right]_3^5 \\ & = {1 \over 25} \left[ \ln (5 + 3) - \ln (5 - 2) - {5 \over 5 - 2} \right] - {1 \over 25} \left[ \ln (3 + 3) - \ln (3 - 2) - {5 \over 3 - 2} \right] \\ & = {1 \over 25} \left( \ln 8 - \ln 3 - {5 \over 3} \right) - {1 \over 25} \left( \ln 6 - \ln 1 - {5 \over 1} \right) \\ & = {1 \over 25} \left( \ln 8 - \ln 3 - {5 \over 3} \right) - {1 \over 25} \left( \ln 6 - 0 - 5 \right) \\ & = {1 \over 25} \left[ \ln 8 - \ln 3 - {5 \over 3} - \left(\ln 6 - 5 \right) \right] \\ & = {1 \over 25} \left( \ln 8 - \ln 3 - {5 \over 3} - \ln 6 + 5 \right) \\ & = {1 \over 25} \left( \ln 8 - \ln 3 - \ln 6 + {10 \over 3} \right) \\ & = {1 \over 25} \left( \ln {8 \over 3} - \ln 6 + {10 \over 3} \right) \phantom{00000000} [\text{Quotient law (logarithms)}] \\ & = {1 \over 25} \left( \ln {{8 \over 3} \over 6} + {10 \over 3} \right) \phantom{000000000000} [\text{Quotient law (logarithms)}] \\ & = {1 \over 25} \left( \ln {4 \over 9} + {10 \over 3} \right) \\ & = {1 \over 25} \left[ \ln \left(2 \over 3\right)^2 + {10 \over 3} \right] \\ & = {1 \over 25} \left[ 2\ln \left(2 \over 3\right) + {10 \over 3} \right] \phantom{0000000000} [\text{Power law (logarithms)}] \end{align}

(i)

\begin{align} {dy \over dx} & = (-\sin x)(e^{\cos x}) \\ & = -(\sin x)e^{\cos x} \end{align}

(ii)

\begin{align} \text{From (i), } {d \over dx}(e^{\cos x}) & = -(\sin x)e^{\cos x} \\ \\ \implies \int -(\sin x)e^{\cos x} \phantom{.} dx & = e^{\cos x} \\ - \int (\sin x)e^{\cos x} \phantom{.} dx & = e^{\cos x} \\ \int (\sin x)e^{\cos x} \phantom{.} dx & = - e^{\cos x} \\ \\ \therefore \int \sin x (e^{\cos x} + 1) \phantom{.} dx & = \int (\sin x) e^{\cos x} + \sin x \phantom{.} dx \\ & = - e^{\cos x} + (-\cos x) \\ & = - e^{\cos x} - \cos x + c \end{align}

(a)

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

(b)

\begin{align} \int (e^{{1 \over 3}x - 1} - 3\cos 2x) \phantom{.} dx & = {e^{{1 \over 3}x - 1} \over {1 \over 3}} - 3 \left({\sin 2x \over 2}\right) \\ & = 3e^{{1 \over 3}x - 1} - {3 \over 2}\sin 2x + c \end{align}

\begin{align} {dy \over dx} & = {2x^2 + 1 \over \sqrt{x}} - \sqrt{e^{-{1 \over 2}x}} \\ & = {2x^2 + 1 \over x^{1 \over 2}} - \left( e^{-{1 \over 2}x} \right)^{1 \over 2} \\ & = {2x^2 \over x^{1 \over 2}} + {1 \over x^{1 \over 2}} - e^{-{1 \over 2}x\left(1 \over 2\right)} \\ & = 2x^{2 - {1 \over 2}} + x^{-{1 \over 2}} - e^{-{1 \over 4}x} \\ & = 2x^{3 \over 2} + x^{-{1 \over 2}} - e^{-{1 \over 4}x} \\ \\ y & = \int {dy \over dx} \phantom{.} dx \\ & = \int 2x^{3 \over 2} + x^{-{1 \over 2}} - e^{-{1 \over 4}x} \phantom{.} dx \\ & = {2x^{5 \over 2} \over {5 \over 2}} + {x^{1 \over 2} \over {1 \over 2}} - {e^{-{1 \over 4}x} \over -{1 \over 4}} \\ & = {4 \over 5}x^{5 \over 2} + 2x^{1 \over 2} + 4e^{-{1 \over 4}x} + c \\ \\ \text{When } & x = 0 \text{ and } y = 2, \\ 2 & = {4 \over 5}(0)^{5 \over 2} + 2(0)^{1 \over 2} + 4e^{-{1 \over 4}(0)} + c \\ 2 & = 0 + 0 + 4e^0 + c \\ 2 & = 4(1) + c \\ 2 & = 4 + c \\ 2 - 4 & = c \\ -2 & = c \\ \\ \therefore y & = {4 \over 5}x^{5 \over 2} + 2x^{1 \over 2} + 4e^{-{1 \over 4}x} - 2 \end{align}

(i)

\begin{align} \require{cancel} \text{Let } y & = \ln (\sec x + \tan x) \\ & = \ln \left( {1 \over \cos x} + {\sin x \over \cos x} \right) \\ & = \ln \left( {1 + \sin x \over \cos x } \right) \\ & = \ln (1 + \sin x) - \ln (\cos x) \phantom{000000} [\text{Quotient law (logarithms)}] \\ \\ {dy \over dx} & = {\cos x \over 1 + \sin x} - {-\sin x \over \cos x} \\ & = {\cos x \over 1 + \sin x} + {\sin x \over \cos x} \\ & = {\cos x(\cos x) \over \cos x(1 + \sin x)} + {\sin x(1 + \sin x) \over \cos x(1 + \sin x)} \\ & = { \cos^2 x \over \cos x(1 + \sin x) } + { \sin x + \sin^2 x \over \cos x(1 + \sin x)} \\ & = { \cos^2 x + \sin x + \sin^2 x \over \cos x(1 + \sin x)} \\ & = { \sin^2 x + \cos^2 x + \sin x \over \cos x(1 + \sin x)} \\ & = { \cancel{1 + \sin x} \over \cos x \cancel{(1 + \sin x)}} \phantom{000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = { 1 \over \cos x} \\ & = \sec x \end{align}

(ii)

\begin{align} \text{From (i), } {d \over dx}& [\ln (\sec x + \tan x)] = \sec x \\ \\ \implies \int \sec x \phantom{.} dx & = \ln (\sec x + \tan x) + c \end{align}

$$ \require{enclose} \begin{array}{rll} x - a \phantom{000000}\\ x + a \enclose{longdiv}{ x^2 + a \phantom{000000}}\kern-.2ex \\ -\underline{( x^2 + ax ){\phantom{0000.}}} \\ -ax + a\phantom{00} \\ -\underline{( - ax - a^2){\phantom{.}}} \\ a + a^2\phantom{0} \end{array} $$
\begin{align} \int {x^2 + a \over x + a} \phantom{.} dx & = \int x - a + {a + a^2 \over x + a} \phantom{.} dx \\ & = \int x - a \phantom{.} dx \phantom{.} + \int {a + a^2 \over x + a} \phantom{.} dx \\ & = {x^2 \over 2} - ax + (a + a^2) \int {1 \over x + a} \phantom{.} dx \\ & = {1 \over 2}x^2 - ax + (a + a^2) \ln (x + a) + c \\ & = {1 \over 2}x^2 - ax + a(1 + a) \ln (x + a) + c \end{align}

\begin{align} \text{For } x \le 0, & \phantom{.} \ln x \text{ is undefined} \\ \\ \therefore \int {1 \over x} \phantom{.} dx & = \ln |x| + c \end{align}

\begin{align} {d \over dx}(e^{-x^2}) & = -2x e^{-x^2} \\ \\ \implies \int -2x e^{-x^2} \phantom{.} dx & = e^{-x^2} \\ -2 \int x e^{-x^2} \phantom{.} dx & = e^{-x^2} \\ \int x e^{-x^2} \phantom{.} dx & = {e^{-x^2} \over 2} + c \\ \\ \text{Student is wrong as the variable } & x \text{ cannot be factorised from the integral} \end{align}