Given that $$\frac{13 - 4x}{(2x + 1)(x + 3)} = \frac{A}{(2x + 1)} + \frac{B}{(2x + 1)^2} + \frac{C}{(x + 3)}$$ (a) find the values of the constants A, B and C - Edexcel - A-Level Maths Pure - Question 5 - 2018 - Paper 9

Using the constants found, we can rewrite the function:

$\frac{13 - 4x}{(2x + 1)(x + 3)} = \frac{-2}{(2x + 1)} + \frac{6}{(2x + 1)^2} + \frac{1}{(x + 3)}$

Now integrate each term separately:

1. Integrate $\frac{-2}{(2x + 1)}$:

   $\int \frac{-2}{(2x + 1)} \, dx = -2 \ln|2x + 1| + C_1$

2. Integrate $\frac{6}{(2x + 1)^2}$:

   $\int \frac{6}{(2x + 1)^2} \, dx = -3(2x + 1)^{-1} + C_2 = -\frac{3}{2x + 1} + C_2$

3. Integrate $\frac{1}{(x + 3)}$:

   $\int \frac{1}{(x + 3)} \, dx = \ln|x + 3| + C_3$

Combining all these gives:

$\int \frac{13 - 4x}{(2x + 1)(x + 3)} \, dx = -2 \ln|2x + 1| - \frac{3}{2x + 1} + \ln|x + 3| + C$

To find this integral, we can separate the terms and integrate:

$\int (e^x + 1) \, dx = \int e^x \, dx + \int 1 \, dx$

The integral of $e^x$ is $e^x$ and the integral of $1$ is $x$. Thus:

$= e^x + x + C$

Using the substitution $u^2 = x$ leads to $dx = 2u \, du$ and we change the limits:

Rewrite the integral:

$\int \frac{1}{4x + 5x^2} \, dx = \int \frac{1}{5u^4 + 4u^2} 2u \, du$

Now factor the denominator:

$\int \frac{2u}{u^2(5u + 4)} \, du = \int \left(\frac{2}{5u + 4} - \frac{8}{5} \right) \, du$

Integrating gives:

$= \frac{2}{5} \ln|5u + 4| - \frac{8}{5}u + C$

Finally, substitute back for $x$:

$= \frac{2}{5} \ln(5\sqrt{x} + 4) - \frac{8}{5}\sqrt{x} + C$