Area bounded between the curve $y=x^2 - 4x$ and $y= 2x/(x-3)$ I've determined the intersects to be $x = 0, 2, 5$ and that $\frac{2x}{x-3}$, denoted as $f(x)$, is above $x(x-4)$, denoted as $g(x)$, so to find the area, I'll need to find the integral from $0$ to $2$ of $f(x) - g(x)$. But I've been stuck for a while playing around with this question.

Notice, the area bounded by the curves from $x=0$ to $x=2$, is given as $$\int_{0}^{2}\left(\frac{2x}{x-3}-(x^2-4x)\right)\ dx$$
$$=\int_{0}^{2}\left(\frac{2(x-3)+6}{x-3}-x^2+4x\right)\ dx$$
$$=\int_{0}^{2}\left(2+\frac{6}{x-3}-x^2+4x\right)\ dx$$
$$=\left(2x+6\ln|x-3|-\frac{x^3}{3}+2x^2\right)_{0}^{2}$$
$$=4-\frac{8}{3}+8-6\ln (3)$$$$=\color{red}{\frac{28}{3}-6\ln (3)}$$
