express the integrand as a sum of partial fractions and evaluate the integral.\n int_{8}^{12}\frac{y}{y^{2}-4…

Explanation:

Step1: Factor the denominator

First, factor (y^{2}-4y - 5=(y - 5)(y+1)). Then, assume (\frac{y}{y^{2}-4y - 5}=\frac{A}{y - 5}+\frac{B}{y + 1}). Cross - multiply to get (y=A(y + 1)+B(y - 5)).

Step2: Find the values of A and B

Let (y = 5), then (5=A(5 + 1)+B(5 - 5)), so (A=\frac{5}{6}). Let (y=-1), then (-1=A(-1 + 1)+B(-1 - 5)), so (B=\frac{1}{6}). Thus, (\frac{y}{y^{2}-4y - 5}=\frac{5/6}{y - 5}+\frac{1/6}{y + 1}).

Step3: Evaluate the integral

(\int_{8}^{12}\frac{y}{y^{2}-4y - 5}dy=\int_{8}^{12}(\frac{5/6}{y - 5}+\frac{1/6}{y + 1})dy=\frac{5}{6}\int_{8}^{12}\frac{1}{y - 5}dy+\frac{1}{6}\int_{8}^{12}\frac{1}{y + 1}dy). Using the integral formula (\int\frac{1}{u}du=\ln|u|+C), we have (\frac{5}{6}[\ln|y - 5|]{8}^{12}+\frac{1}{6}[\ln|y + 1|]{8}^{12}). (\frac{5}{6}(\ln(12 - 5)-\ln(8 - 5))+\frac{1}{6}(\ln(12 + 1)-\ln(8 + 1))=\frac{5}{6}(\ln7-\ln3)+\frac{1}{6}(\ln13-\ln9)). Using the property (\ln a-\ln b=\ln\frac{a}{b}), we get (\frac{5}{6}\ln\frac{7}{3}+\frac{1}{6}\ln\frac{13}{9}=\frac{1}{6}(5\ln\frac{7}{3}+\ln\frac{13}{9})=\frac{1}{6}(\ln(\frac{7}{3})^{5}+\ln\frac{13}{9})=\frac{1}{6}\ln(\frac{7^{5}}{3^{5}}\cdot\frac{13}{9})=\frac{1}{6}\ln(\frac{16807\times13}{243\times9})=\frac{1}{6}\ln(\frac{218491}{2187})).

Answer:

(\frac{y}{y^{2}-4y - 5}=\frac{5/6}{y - 5}+\frac{1/6}{y + 1}); (\int_{8}^{12}\frac{y}{y^{2}-4y - 5}dy=\frac{1}{6}\ln(\frac{218491}{2187}))