Find \( \int \frac{x}{9 - 4x^2} \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 1 - 2006 - Paper 1

Completing the square gives: ( x^2 - 6x + 13 = (x - 3)^2 + 4 ).

Thus, the integral becomes:

$\int \frac{dx}{(x - 3)^2 + 4}$

Using the substitution ( v = x - 3 ), the integral transforms to:

$\int \frac{dv}{v^2 + 4} = \frac{1}{2} \tan^{-1} \left( \frac{v}{2} \right) + C = \frac{1}{2} \tan^{-1} \left( \frac{x - 3}{2} \right) + C$

To find a, b, and c, set up the equation:

$16x - 43 = a(x + 2) + b(x - 3) + c$

Expanding and collecting terms:

$16x - 43 = (a + b)x + (2a - 3b + c)$

Equating coefficients:

1. ( a + b = 16 )
2. ( 2a - 3b + c = -43 )

Choosing values for a, b, and c, one solution can be found: ( a = 10, b = 6, c = -17 ).

Now, integrate term by term:

$\int \frac{10}{x - 3} \, dx + \int \frac{6}{x + 2} \, dx - \int \frac{17}{(x - 3)(x + 2)} \, dx$

Calculating this results in:

$10 \ln |x - 3| + 6 \ln |x + 2| - 17 \left( \ln |(x - 3)| - \ln |(x + 2)| \right) + C$

Using integration by parts, let ( u = t ) and ( dv = e^{-t} dt ), then ( du = dt ) and ( v = -e^{-t} ):

$\int te^{-t} \, dt = -te^{-t} - \int -e^{-t} \, dt = -te^{-t} + e^{-t} + C$

Evaluating from 0 to 2:

$\left[-2e^{-2} + e^{-2} \right] - [0 + 1] = -e^{-2} - 1$

Using the substitution, we find:

$\sin \theta = \frac{2t}{1 + t^2}, \quad d\theta = \frac{2 \, dt}{1 + t^2}\n\int \frac{d\theta}{\sin \theta} = \int \frac{2 \, dt}{2t} = \int \frac{dt}{t} = \ln |t| + C$

From here, evaluate the limits, leading to the conclusion ( \frac{1}{2} \log 3 ).