Finding the area under $\frac{3}{2x+1}$ and $3x-2$ Find the area between these curves.
$$y=\dfrac{3}{2x+1},\qquad y=3x-2;\qquad x=2\quad  \text{et} \quad y=0 $$

Indeed,
I calculate the integral of the blue function between $1$ and $2$. Then, I will calculate the area of the triangle between the yellow line and the x-axis on the interval $\dfrac{2}{3}$ and $1$, which gives me:
$$\int _{\frac{3}{2}}^1\left(3x-2\right)dx+\int _1^2\frac{3}{2x+1}dx=\frac{3}{2}\ln \left(\frac{5}{3}\right)-\frac{7}{8}$$
Am I right in the way how to find calculate the area and if that so that $\frac{3}{2}\ln \left(\frac{5}{3}\right)-\frac{7}{8}$ is it correct?

i think the first integral must be $$\int_{2/3}^{1}(3x-2)dx$$
