Properties:

\begin{align*} \text{Area above } x \text{-axis} & = \int_0^{ {7 \over 18} \pi} 2 \sin 3x + 1 \phantom{.} dx \\ & = \left[ {-2 \cos 3x \over 3} + x \right]_0^{ {7 \over 18} \pi} \phantom{000000} \left[ \int \sin f(x) \phantom{.} dx = { - \cos f(x) \over f'(x)} \right] \\ & = \left[ -{2 \over 3} \cos 3x + x \right]_0^{ {7 \over 18} \pi} \\ & = -{2 \over 3} \cos {7 \over 6}\pi + {7 \over 18} \pi - \left( -{2 \over 3} \cos 0 + 0 \right) \\ & = -{2 \over 3} \left(-{\sqrt{3} \over 2}\right) + {7 \over 18} \pi + {2 \over 3} (1) \\ & = {\sqrt{3} \over 3} + {7 \over 18} \pi + {2 \over 3} \\ \\ \text{Area below } x \text{-axis} & = -\int_{ {7 \over 18} \pi}^{ {11 \over 18} \pi} 2 \sin 3x + 1 \phantom{.} dx \\ & = -\left[ -{2 \over 3} \cos 3x + x \right]_{ {7 \over 18} \pi}^{ {11 \over 18} \pi} \phantom{0000000} [\text{Same integration as above}] \\ & = - \left[ -{2 \over 3} \cos {11 \over 6} \pi + {11 \over 18} \pi - \left( -{2 \over 3} \cos {7 \over 6}\pi + {7 \over 18} \pi \right) \right] \\ & = - \left[ -{2 \over 3} \left(\sqrt{3} \over 2\right) + {11 \over 18} \pi + {2 \over 3} \left(-\sqrt{3} \over 2\right) - {7 \over 18} \pi \right] \\ & = {\sqrt{3} \over 3} - {11 \over 18} \pi + {\sqrt{3} \over 3} + {7 \over 18} \pi \\ & = {2 \sqrt{3} \over 3} - {4 \over 18} \pi \\ \\ \text{Total area} & = {\sqrt{3} \over 3} + {7 \over 18} \pi + {2 \over 3} + {2 \sqrt{3} \over 3} - {4 \over 18} \pi \\ & = {3 \sqrt{3} \over 3} + {3 \over 18} \pi + {2 \over 3} \\ & = \left( \sqrt{3} + {3 \over 18} \pi + {2 \over 3} \right) \text{ units}^2 \end{align*}

\begin{align*} y & = \ln (x + 1) \\ y & = \log_e (x + 1) \\ e^y & = x + 1 \\ \\ \therefore x & = e^y - 1 \end{align*}

\begin{align*} \text{Area} & = \int_0^{\ln 4} e^y - 1 \phantom{.} dy \phantom{0000000.} [\text{Integrate with respect to } y] \\ & = \left[ {e^y \over 1} - y \right]_0^{\ln 4} \phantom{000000000} \left[ \int e^{f(x)} \phantom{.} dx = {e^{f(x)} \over f'(x)} \right] \\ & = e^{\ln 4} - \ln 4 - (e^0 - 0) \\ & = 4 - \ln 4 - 1 \\ & = (3 - \ln 4) \text{ units}^2 \end{align*}

\begin{align*} {3x + 1 \over x - 1} & = {a \over 1} + {b \over x - 1} \\ & = {a(x - 1) \over x - 1} + {b \over x - 1} \\ & = {a(x - 1) + b \over x - 1} \\ & = {ax - a + b \over x - 1} \\ \\ 3x + 1 & = ax - a + b \\ \\ \text{Comparing } & \text{coefficients of } x , \\ a & = 3 \\ \\ \text{Comparing } & \text{constants,} \\ 1 & = - a + b \\ 1 & = - 3 + b \\ 4 & = b \\ \\ \therefore {3x + 1 \over x - 1} & = 3 + {4 \over x - 1} \end{align*}