Express $\frac{8x+13}{(1+2x)(2+x)^2}$ in partial fractions. Express $\frac{8x+13}{(1+2x)(2+x)^2}$ in partial fractions. Hence, evaluate $\displaystyle \int_{1}^{2} \frac{8x+13}{(1+2x)(2+x)^2}\,\mathrm{d}x$.

1. Set up the partial fraction decomposition form

Let $\frac{8x+13}{(1+2x)(2+x)^2} = \frac{A}{1+2x} + \frac{B}{2+x} + \frac{C}{(2+x)^2}$

2. Form the basic equation by clearing the denominator

$8x+13 = A(2+x)^2 + B(1+2x)(2+x) + C(1+2x)$

3. Find A by substituting x = -1/2

Substitute $x = -\frac{1}{2}$:
$8\left(-\frac{1}{2}\right) + 13 = A\left(2 - \frac{1}{2}\right)^2$
$-4 + 13 = A\left(\frac{3}{2}\right)^2$
$9 = \frac{9}{4}A \implies A = 4 \times \frac{9}{9} = 4$

4. Find C by substituting x = -2

Substitute $x = -2$:
$8(-2) + 13 = C(1 + 2(-2))$
$-16 + 13 = C(1 - 4)$
$-3 = -3C \implies C = 1$

5. Find B by substituting x = 0 (or equating coefficients)

Substitute $x = 0$, $A = 4$, $C = 1$:
$13 = 4(2)^2 + B(1)(2) + 1(1)$
$13 = 16 + 2B + 1$
$13 = 17 + 2B$
$2B = -4 \implies B = -2$

Therefore, $\frac{8x+13}{(1+2x)(2+x)^2} = \frac{4}{1+2x} - \frac{2}{2+x} + \frac{1}{(2+x)^2}$

1. Find the partial fractions from part (a)

Let $\frac{8x+13}{(1+2x)(2+x)^2} = \frac{A}{1+2x} + \frac{B}{2+x} + \frac{C}{(2+x)^2}$
$8x+13 = A(2+x)^2 + B(1+2x)(2+x) + C(1+2x)$
Let $x = -2$: $-16+13 = C(1-4) \Rightarrow -3 = -3C \Rightarrow C = 1$
Let $x = -\frac{1}{2}$: $-4+13 = A(2-\frac{1}{2})^2 \Rightarrow 9 = A(\frac{9}{4}) \Rightarrow A = 4$
Compare coefficients of $x^2$: $0 = A + 2B \Rightarrow 0 = 4 + 2B \Rightarrow B = -2$
So, $\frac{8x+13}{(1+2x)(2+x)^2} = \frac{4}{1+2x} - \frac{2}{2+x} + \frac{1}{(2+x)^2}$

2. Integrate the partial fractions

$\int_{1}^{2} \frac{8x+13}{(1+2x)(2+x)^2}\,\mathrm{d}x = \int_{1}^{2} \left( \frac{4}{1+2x} - \frac{2}{2+x} + (2+x)^{-2} \right) \,\mathrm{d}x$
$= \left[ \frac{4\ln(1+2x)}{2} - 2\ln(2+x) + \frac{(2+x)^{-1}}{-1} \right]_{1}^{2}$
$= \left[ 2\ln(1+2x) - 2\ln(2+x) - \frac{1}{2+x} \right]_{1}^{2}$

3. Evaluate the definite integral

$= \left( 2\ln 5 - 2\ln 4 - \frac{1}{4} \right) - \left( 2\ln 3 - 2\ln 3 - \frac{1}{3} \right)$
$= 2\ln 5 - 2\ln 4 - \frac{1}{4} + \frac{1}{3}$
$= 2\ln\left(\frac{5}{4}\right) + \frac{1}{12}$
$= \ln\left(\frac{25}{16}\right) + \frac{1}{12}$