Using standard series from the List of Formula (MF26), expand $\ln(k + 2x)^n$ in ascending powers of $x$, as far as the term in $x^3$, where $k$ and $n$ are real constants, $k > 0$, $n < 0$ and $x > -\frac{k}{2}$. State the range of values of $x$ for which the expansion in part (i) is valid. Find the exact value of $\int_{0}^{1} \ln(1 + 2x)^{-2} \; \mathrm{d}x$. Use your series from part (i) to estimate $\int_{0}^{1} \ln(1 + 2x)^{-2} \; \mathrm{d}x$. Calculate the percentage error of the estimate in part (iv), and comment on the accuracy of your approximation using your answer from part (ii).

1. Use logarithm properties to rewrite the expression into a form suitable for the standard series expansion.

$\ln(k + 2x)^n = n\ln(k + 2x)$
$= n\ln\left[k\left(1 + \frac{2x}{k}\right)\right]$
$= n\ln k + n\ln\left(1 + \frac{2x}{k}\right)$

2. Apply the standard Maclaurin series expansion for $\ln(1+x)$ from the MF26 formula list.

From MF26, the standard series is $\ln(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \dots$
Substitute $y = \frac{2x}{k}$:
$\ln\left(1 + \frac{2x}{k}\right) = \left(\frac{2x}{k}\right) - \frac{1}{2}\left(\frac{2x}{k}\right)^2 + \frac{1}{3}\left(\frac{2x}{k}\right)^3 - \dots$
$= \frac{2x}{k} - \frac{2x^2}{k^2} + \frac{8x^3}{3k^3} - \dots$

3. Multiply by $n$ and simplify to obtain the final expansion up to the term in $x^3$.

$n\ln(k + 2x) = n\ln k + n\left( \frac{2x}{k} - \frac{2x^2}{k^2} + \frac{8x^3}{3k^3} - \dots \right)$
$= n\ln k + \frac{2n}{k}x - \frac{2n}{k^2}x^2 + \frac{8n}{3k^3}x^3 + \dots$

1. State the condition for the expansion of the logarithmic term to be valid

The expansion in part (i) relies on the standard series for $\ln(1 + y)$ where $y = \frac{2x}{k}$.
The standard expansion for $\ln(1 + y)$ is valid for $-1 < y \le 1$.
Therefore, we require:
$-1 < \frac{2x}{k} \le 1$

2. Solve the inequality for $x$

Since it is given that $k > 0$, we can safely multiply the inequality by $\frac{k}{2}$ without flipping the inequality signs:
$-\frac{k}{2} < x \le \frac{k}{2}$

1. Simplify the integrand using logarithm properties and set up integration by parts.

$\int_{0}^{1} \ln(1 + 2x)^{-2} \; \mathrm{d}x = -2 \int_{0}^{1} \ln(1 + 2x) \; \mathrm{d}x$
Let $u = \ln(1+2x)$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = 1$:
$\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2}{1+2x}, \quad v = x$

2. Apply the integration by parts formula to the definite integral.

Using $\int u \frac{\mathrm{d}v}{\mathrm{d}x} \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \mathrm{d}x$:
$-2 \int_{0}^{1} \ln(1 + 2x) \; \mathrm{d}x = -2 \left[ x \ln(1+2x) \right]_0^1 - (-2) \int_{0}^{1} x \left(\frac{2}{1+2x}\right) \; \mathrm{d}x$
$= -2(1 \cdot \ln 3 - 0) + \int_{0}^{1} \frac{4x}{1+2x} \; \mathrm{d}x$
$= -2\ln 3 + \int_{0}^{1} \frac{4x}{1+2x} \; \mathrm{d}x$

3. Use algebraic manipulation to integrate the rational fraction.

$\frac{4x}{1+2x} = \frac{2(1+2x) - 2}{1+2x} = 2 - \frac{2}{1+2x}$
$\int_{0}^{1} \left( 2 - \frac{2}{1+2x} \right) \; \mathrm{d}x = \left[ 2x - \ln(1+2x) \right]_0^1$
$= (2 - \ln 3) - (0 - \ln 1)$
$= 2 - \ln 3$

4. Combine the results to obtain the final exact value.

$\text{Exact value} = -2\ln 3 + 2 - \ln 3 = 2 - 3\ln 3$

1. Derive the series expansion for $\ln(k+2x)^n$ required from part (i).

$\text{As required from (i):}$
$\ln(k + 2x)^n = n \ln\left[k\left(1 + \frac{2x}{k}\right)\right] = n \ln k + n\ln\left(1 + \frac{2x}{k}\right)$
$\approx n \ln k + n \left( \frac{2x}{k} - \frac{1}{2}\left(\frac{2x}{k}\right)^2 + \frac{1}{3}\left(\frac{2x}{k}\right)^3 \right)$
$= n \ln k + \frac{2n}{k}x - \frac{2n}{k^2}x^2 + \frac{8n}{3k^3}x^3$

2. Substitute $k=1$ and $n=-2$ into the series expansion to approximate $\ln(1+2x)^{-2}$.

$\text{Matching } \ln(1 + 2x)^{-2} \text{ with } \ln(k + 2x)^n, \text{ let } k = 1 \text{ and } n = -2:$
$\ln(1 + 2x)^{-2} \approx -2 \ln 1 + \frac{2(-2)}{1}x - \frac{2(-2)}{1^2}x^2 + \frac{8(-2)}{3(1^3)}x^3$
$= -4x + 4x^2 - \frac{16}{3}x^3$

3. Evaluate the definite integral of the polynomial approximation from $x=0$ to $x=1$.

$\int_{0}^{1} \ln(1 + 2x)^{-2} \; \mathrm{d}x \approx \int_{0}^{1} \left( -4x + 4x^2 - \frac{16}{3}x^3 \right) \; \mathrm{d}x$
$= \left[ -2x^2 + \frac{4}{3}x^3 - \frac{4}{3}x^4 \right]_{0}^{1}$
$= \left(-2(1)^2 + \frac{4}{3}(1)^3 - \frac{4}{3}(1)^4\right) - (0)$
$= -2$

1. Calculate the exact value and estimated value from parts (iii) and (iv) to be used for the percentage error.

From part (iii), the exact value is:
$\int_{0}^{1} \ln(1 + 2x)^{-2} \; \mathrm{d}x = -2\int_{0}^{1} \ln(1 + 2x) \; \mathrm{d}x$
Using the substitution $u = 1+2x$, $\mathrm{d}u = 2\mathrm{d}x$:
$-\int_{1}^{3} \ln u \; \mathrm{d}u = -[u\ln u - u]_1^3 = 2 - 3\ln 3 \approx -1.2958$

From part (iv), using the series expansion for $k=1, n=-2$:
$\ln(1+2x)^{-2} \approx -4x + 4x^2 - \frac{16}{3}x^3$
$\int_{0}^{1} \left(-4x + 4x^2 - \frac{16}{3}x^3\right) \; \mathrm{d}x = \left[-2x^2 + \frac{4}{3}x^3 - \frac{4}{3}x^4\right]_0^1 = -2$

2. Calculate the percentage error of the estimate.

$\text{Percentage error} = \left| \frac{\text{Approximate} - \text{Exact}}{\text{Exact}} \right| \times 100\%$
$= \left| \frac{-2 - (2 - 3\ln 3)}{2 - 3\ln 3} \right| \times 100\%$
$= \left| \frac{3\ln 3 - 4}{2 - 3\ln 3} \right| \times 100\%$
$\approx \left| \frac{-0.70416}{-1.29584} \right| \times 100\% \approx 54.3\%$

3. Comment on the accuracy using the range of validity from part (ii).

In part (ii), the expansion of $\ln(k+2x)^n = n\ln k + n\ln\left(1+\frac{2x}{k}\right)$ is valid for $-1 < \frac{2x}{k} \le 1 \implies -\frac{k}{2} < x \le \frac{k}{2}$.

For $k=1$, the series is valid only for $-\frac{1}{2} < x \le \frac{1}{2}$.
Since the integration evaluates from $x=0$ to $x=1$, a portion of the interval $[0, 1]$ lies outside the range of validity for the series. Therefore, the series diverges or is grossly inaccurate for $x > 0.5$, resulting in a poor overall approximation (as evidenced by the large $54.3\%$ error).