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.
 A: For Area between 0 to 2
$$
Area = \int_0^2 [(\frac{2x}{x-3}) - (x^{2}-4x)]dx\
$$
A: First, simplify the difference between the two functions:
$$
\begin{eqnarray}
f(x) - g(x) &=& \frac{2x}{x-3} - x(x - 4) \\
&=& \frac{-x^3 + 7x^2 - 10x}{x-3} \\
\end{eqnarray}
$$
Then, integrate by substituting $u = x-3$:
$$
\begin{eqnarray}
\int_0^2 \! \frac{-x^3 + 7x^2 - 10x}{x-3} \, \textrm{d}x &=& \int_{-3}^{-1} \! \frac{-(u+3)^3 + 7(u+3)^2 - 10(u+3)}{u} \, \textrm{d}u \\
&=& \int_{-3}^{-1} \! \frac{-u^3 - 2u^2 + 5u + 6}{u} \, \textrm{d}u \\
&=& -\frac{u^3}{3} - u^2 + 5u + 6\ln(u) \bigg|_{-3}^{-1} \\
&=& \frac{28}{3} - 6\ln(3) \\
\end{eqnarray}
$$
A: Notice, the area bounded by the curves from $x=0$ to $x=2$, is given as $$\int_{0}^{2}\left(\frac{2x}{x-3}-(x^2-4x)\right)\ dx$$
$$=\int_{0}^{2}\left(\frac{2(x-3)+6}{x-3}-x^2+4x\right)\ dx$$
$$=\int_{0}^{2}\left(2+\frac{6}{x-3}-x^2+4x\right)\ dx$$
$$=\left(2x+6\ln|x-3|-\frac{x^3}{3}+2x^2\right)_{0}^{2}$$
$$=4-\frac{8}{3}+8-6\ln (3)$$$$=\color{red}{\frac{28}{3}-6\ln (3)}$$
