Express $\frac{10x}{3-5x}$ in the form $a + \frac{b}{3-5x}$, where $a$ and $b$ are integers. Hence, integrate $\frac{10x}{3-5x}$ with respect to $x$.

1. Rewrite the numerator in terms of the denominator

We want to express $10x$ in terms of $(3-5x)$.
$10x = -2(-5x) = -2(3 - 5x - 3) = -2(3 - 5x) + 6$

2. Split the fraction and simplify to find $a$ and $b$

$\frac{10x}{3-5x} = \frac{-2(3-5x) + 6}{3-5x}$
$= -2 + \frac{6}{3-5x}$
Thus, $a = -2$ and $b = 6$.

1. Express the integrand using the result from part (a)

From part (a), we find $a$ and $b$ such that $\frac{10x}{3-5x} = a + \frac{b}{3-5x}$.
$10x = a(3-5x) + b = -5ax + (3a+b)$
Comparing coefficients:
$-5a = 10 \implies a = -2$
$3a + b = 0 \implies 3(-2) + b = 0 \implies b = 6$
So, $\frac{10x}{3-5x} = -2 + \frac{6}{3-5x}$.

2. Integrate the expression term by term

$\int \frac{10x}{3-5x} \, dx = \int \left( -2 + \frac{6}{3-5x} \right) \, dx$
$= -2x + \frac{6}{-5} \ln(3-5x) + c$
$= -2x - \frac{6}{5} \ln(3-5x) + c$