Express $\frac{2x+15}{(3+x)(3-2x)}$ in partial fractions. Hence evaluate $\int_{0}^{1} \frac{2x+15}{(3+x)(3-2x)} \, dx$.

1. Set up the partial fraction decomposition with unknown constants

Let $\frac{2x+15}{(3+x)(3-2x)} = \frac{A}{3+x} + \frac{B}{3-2x}$

2. Form the basic algebraic equation

Multiply throughout by $(3+x)(3-2x)$:
$2x+15 = A(3-2x) + B(3+x)$

3. Solve for the constants by substituting suitable values of $x$

Substitute $x = -3$:
$2(-3) + 15 = A(3 - 2(-3))$
$9 = 9A \implies A = 1$

Substitute $x = \frac{3}{2}$:
$2\left(\frac{3}{2}\right) + 15 = B\left(3 + \frac{3}{2}\right)$
$18 = \frac{9}{2}B \implies B = 4$

4. State the final partial fraction expression

$\frac{2x+15}{(3+x)(3-2x)} = \frac{1}{3+x} + \frac{4}{3-2x}$

1. Express the integrand in partial fractions (from part i)

Let $\frac{2x+15}{(3+x)(3-2x)} = \frac{A}{3+x} + \frac{B}{3-2x}$.
$2x+15 = A(3-2x) + B(3+x)$
Substitute $x = -3$: $9 = 9A \implies A = 1$
Substitute $x = 1.5$: $18 = 4.5B \implies B = 4$
$\frac{2x+15}{(3+x)(3-2x)} = \frac{1}{3+x} + \frac{4}{3-2x}$

2. Integrate the partial fractions

$\int_{0}^{1} \left( \frac{1}{3+x} + \frac{4}{3-2x} \right) \, dx$
$= \left[ \ln(3+x) + \frac{4}{-2}\ln(3-2x) \right]_{0}^{1}$
$= \left[ \ln(3+x) - 2\ln(3-2x) \right]_{0}^{1}$

3. Evaluate the definite integral by substituting the limits

Substitute $x = 1$:
$\ln(3+1) - 2\ln(3-2(1)) = \ln 4 - 2\ln 1 = \ln 4$

Substitute $x = 0$:
$\ln(3+0) - 2\ln(3-2(0)) = \ln 3 - 2\ln 3 = -\ln 3$

Evaluate:
$(\ln 4) - (-\ln 3) = \ln 4 + \ln 3$

4. Simplify the final answer using laws of logarithms

$\ln 4 + \ln 3 = \ln(4 \times 3) = \ln 12$