express the given integrand as a sum of partial fractions, then evaluate the integrals.\n int_{1/4}^{1}…

Explanation:

Step1: Factor the denominator

First, factor out $\frac{1}{4}y$ from the denominator $\frac{1}{4}y^{2}+3y$. We get $\frac{1}{4}y(y + 12)$. So the integrand is $\frac{4y + 15}{\frac{1}{4}y(y + 12)}=\frac{16y+60}{y(y + 12)}$. Let $\frac{16y + 60}{y(y + 12)}=\frac{A}{y}+\frac{B}{y + 12}$. Then $16y+60=A(y + 12)+By=(A + B)y+12A$.

Step2: Solve for A and B

Set up the system of equations: $\begin{cases}A + B=16\12A=60\end{cases}$. From $12A = 60$, we find $A = 5$. Substitute $A = 5$ into $A + B=16$, we get $5 + B=16$, so $B = 11$. So $\frac{16y+60}{y(y + 12)}=\frac{5}{y}+\frac{11}{y + 12}$.

Step3: Integrate the partial - fractions

$\int\frac{4y + 15}{\frac{1}{4}y^{2}+3y}dy=\int(\frac{5}{y}+\frac{11}{y + 12})dy$. Using the integral formula $\int\frac{1}{u}du=\ln|u|+C$, we have $\int(\frac{5}{y}+\frac{11}{y + 12})dy=5\int\frac{1}{y}dy+11\int\frac{1}{y + 12}dy$. $=5\ln|y|+11\ln|y + 12|+C$.

Answer:

$5\ln|y|+11\ln|y + 12|+C$