Consider the definite integral $S = \int _ { 0 } ^ { a } x \sqrt { \frac { 1 } { 3 } x + 2 } \, d x$.
(1) Set $t = \sqrt { \frac { 1 } { 3 } x + 2 }$. Then we have
$$\begin{aligned} \int x \sqrt { \frac { 1 } { 3 } x + 2 } \, d x & = \mathbf { N O } \int \left( t ^ { \mathbf { P } } - \mathbf { Q } t ^ { \mathbf { R } } \right) d t \\ & = \mathbf { S } + C , \end{aligned}$$
where $C$ is the integral constant.
(2) Using the result in (1), we have
$$S = \mathbf { T } .$$
Thus we obtain
$$\lim _ { a \rightarrow \infty } \frac { S } { a ^ { \frac { \mathbf { U } } { \mathbf{V} } } } = \frac { \mathbf { W } \sqrt { \mathbf { X } } } { \mathbf { Y Z } }$$
For $\mathbf{S}$ and $\mathbf{T}$, choose the appropriate expression from among the choices (0) $\sim$ (9) below.
(0) $\frac { 6 } { 5 } t ^ { 5 } \left( 3 t ^ { 2 } - 10 \right)$
(1) $\frac { 6 } { 5 } t ^ { 3 } \left( 3 t ^ { 2 } - 10 \right)$
(2) $\frac { 12 } { 5 } t ^ { 5 } \left( 3 t ^ { 2 } - 5 \right)$
(3) $\frac { 12 } { 5 } t ^ { 3 } \left( 3 t ^ { 2 } - 5 \right)$
(4) $\frac { 6 } { 5 } t ^ { 3 } \left( 3 t ^ { 2 } - 5 \right)$
(5) $\frac { 6 } { 5 } \left\{ \left( \sqrt { \frac { 1 } { 3 } a + 2 } \right) ^ { 5 } ( a - 4 ) + 8 \sqrt { 2 } \right\}$
(6) $\frac { 12 } { 5 } \left\{ \left( \sqrt { \frac { 1 } { 3 } a + 2 } \right) ^ { 3 } ( a - 2 ) + 4 \sqrt { 2 } \right\}$
(7) $\frac { 12 } { 5 } \left\{ \left( \sqrt { \frac { 1 } { 3 } a + 2 } \right) ^ { 5 } ( a - 2 ) + 4 \sqrt { 2 } \right\}$
(8) $\frac { 6 } { 5 } \left\{ \left( \sqrt { \frac { 1 } { 3 } a + 2 } \right) ^ { 3 } ( a - 4 ) + 8 \sqrt { 2 } \right\}$
(9) $\frac { 6 } { 5 } \left\{ \left( \sqrt { \frac { 1 } { 3 } a + 2 } \right) ^ { 3 } ( a - 2 ) + 8 \sqrt { 2 } \right\}$

Consider the function $$f(x) = \frac{\sin x}{3 - 2\cos x} \quad (0 \leqq x \leqq \pi)$$
(1) The derivative of $f(x)$ is $$f'(x) = \frac{\mathbf{A}\cos x - \mathbf{B}}{(\mathbf{C} - \mathbf{D}\cos x)^2}.$$ Let $\alpha$ be the value of $x$ at which $f(x)$ has a local extremum. Then we have $$\cos\alpha = \frac{\mathbf{E}}{\mathbf{F}}.$$
(2) The portion of the plane bounded by the graph of the function $y = f(x)$ and the $x$-axis is divided into two parts by the straight line $x = \alpha$. Let $S_1$ be the area of the part located on the left side of the line. Then we have $$S_1 = \int_{\frac{\mathbf{G}}{\mathbf{H}}}^{\mathbf{I}} \frac{dt}{\mathbf{J} - \mathbf{K}t} = \frac{\mathbf{L}}{\mathbf{L}}\log\frac{\mathbf{L}}{\mathbf{L}}.$$ Let $S_2$ be the area of the part located on the right side. We have $$S_2 = \frac{\mathbf{P}}{2}\log\mathbf{Q}.$$

In the integral $\int \frac { \ln \sqrt { x } } { \sqrt { x } } d x$, if the substitution $u = \sqrt { x }$ is made, which of the following integrals is obtained?
A) $\int \ln u \, d u$
B) $\int 2 \ln u \, d u$
C) $\int \frac { \ln u } { u } d u$
D) $\int \frac { \ln u } { 2 u } d u$
E) $\int u \ln u \, d u$

$$\int _ { 4 } ^ { 9 } \frac { \sqrt { x } } { x - 1 } d x$$
If the substitution $\mathbf { u } = \sqrt { \mathbf { x } }$ is made in the integral, which of the following integrals is obtained?
A) $\int _ { 4 } ^ { 9 } \frac { u } { u ^ { 2 } - 1 } d u$
B) (missing option)
C) $\int _ { 2 } ^ { 3 } \frac { u } { 2 \left( u ^ { 2 } - 1 \right) } d u$
D) $\int _ { 2 } ^ { 3 } \frac { 2 u ^ { 2 } } { u ^ { 2 } - 1 } d u$
E) $\int _ { 2 } ^ { 3 } \frac { u } { u ^ { 2 } - 1 } d u$

$\int \sqrt { 1 + e^{x^{2}} } \, d x$\ In the integral, if the substitution $u = \sqrt { 1 + e ^ { x } }$ is made, which of the following integrals is obtained?\ A) $\int \frac { 2 u } { u ^ { 2 } + 1 } d u$\ B) $\int \frac { u ^ { 2 } } { u ^ { 2 } + 1 } d u$\ C) $\int \frac { 1 } { u ^ { 2 } - 1 } d u$\ D) $\int \frac { u } { u ^ { 2 } - 1 } d u$\ E) $\int \frac { 2 u ^ { 2 } } { u ^ { 2 } - 1 } d u$

$$\int \frac { ( 3 \sqrt { x } + 2 ) ^ { 5 } } { \sqrt { x } } d x$$
Which of the following is this integral equal to? (c is an arbitrary constant.)
A) $\frac { 1 } { 18 } \cdot ( 3 \sqrt { x } + 2 ) ^ { 6 } + c$
B) $\frac { 1 } { 9 } \cdot ( 3 \sqrt { x } + 2 ) ^ { 6 } + c$
C) $\frac { 2 } { 9 } \cdot ( 3 \sqrt { x } + 2 ) ^ { 6 } + c$
D) $\frac { 1 } { 3 } \cdot ( 3 \sqrt { x } + 2 ) ^ { 6 } + c$
E) $\frac { 2 } { 3 } \cdot ( 3 \sqrt { x } + 2 ) ^ { 6 } + c$