Questions: Evaluate the integral. [ int1^2 frac4 y^2-6 y-12y(y+2)(y-3) d y ]

Solution Steps

To evaluate the integral, we can use partial fraction decomposition to break down the integrand into simpler fractions that are easier to integrate. Once we have the partial fractions, we can integrate each term separately.

To evaluate the integral $$\int_{1}^{2} \frac{4 y^{2}-6 y-12}{y(y+2)(y-3)} \, dy,$$ we first decompose the integrand into partial fractions: $$\frac{4 y^{2}-6 y-12}{y(y+2)(y-3)} = \frac{8}{5(y+2)} + \frac{2}{5(y-3)} + \frac{2}{y}.$$

Next, we integrate each term separately: $$\int_{1}^{2} \left( \frac{8}{5(y+2)} + \frac{2}{5(y-3)} + \frac{2}{y} \right) \, dy.$$

We compute the integrals of each term: $$\int_{1}^{2} \frac{8}{5(y+2)} \, dy = \frac{8}{5} \left[ \ln|y+2| \right]_{1}^{2},$$ $$\int_{1}^{2} \frac{2}{5(y-3)} \, dy = \frac{2}{5} \left[ \ln|y-3| \right]_{1}^{2},$$ $$\int_{1}^{2} \frac{2}{y} \, dy = 2 \left[ \ln|y| \right]_{1}^{2}.$$

Evaluating these integrals, we get: $$\frac{8}{5} \left( \ln|4| - \ln|3| \right) = \frac{8}{5} \ln\left(\frac{4}{3}\right),$$ $$\frac{2}{5} \left( \ln|1| - \ln|2| \right) = -\frac{2}{5} \ln(2),$$ $$2 \left( \ln|2| - \ln|1| \right) = 2 \ln(2).$$

Combining these results, we have: $$\frac{8}{5} \ln\left(\frac{4}{3}\right) - \frac{2}{5} \ln(2) + 2 \ln(2).$$

Simplifying further: $$\frac{8}{5} \ln\left(\frac{4}{3}\right) + \left(2 - \frac{2}{5}\right) \ln(2) = \frac{8}{5} \ln\left(\frac{4}{3}\right) + \frac{8}{5} \ln(2).$$

Combining the logarithms: $$\frac{8}{5} \left( \ln\left(\frac{4}{3}\right) + \ln(2) \right) = \frac{8}{5} \ln\left(\frac{4 \cdot 2}{3}\right) = \frac{8}{5} \ln\left(\frac{8}{3}\right).$$

Final Answer