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

Using the values of A, B, and C found in part (a), we can rewrite the integral:

We can integrate each term independently:

1. Integral of A/(2x + 1):

   • Result: $A \ln|2x + 1| + C_1$

2. Integral of B/(2x + 1)^2:

   • Result: $-\frac{B}{2(2x + 1)} + C_2$

3. Integral of C/(x + 3):

   • Result: $C \ln|x + 3| + C_3$

Combining these results, we have:

The final answer will depend on the specific values of A, B, and C.

To solve this integral, we can split it into two parts:

1. First integral: $\int e^{x} \, dx = e^{x} + C_1$
2. Second integral: $\int 1 \, dx = x + C_2$

Thus, combining these results gives:

Using substitution $u^2 = x$, then $dx = 2u \, du$. Thus we can rewrite the integral:

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

Factoring out $u^2$ gives:

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

This can be integrated using partial fractions. Let:

$\frac{1}{u^2(5u^2 + 4)} = \frac{A}{u} + \frac{B}{u^2} + \frac{Cu + D}{5u^2 + 4}$

Solving for A, B, C, and D will give us the form needed to integrate.

The final result corresponds to the appropriate antiderivatives based on the solved constants.