Question

express the given integrand as a sum of partial fractions, then evaluate the integrals.
int_{1/2}^{1} \frac{y + 2}{y^{2}+y}dy
int_{1/2}^{1} \frac{y + 2}{y^{2}+y}dy=square \text{(type an exact answer.)}

Explanation:

Step1: Factor the denominator

First, factor the denominator $\frac{1}{2}y^{2}+y=\frac{1}{2}y(y + 2)$. Let $\frac{y + 2}{\frac{1}{2}y^{2}+y}=\frac{y + 2}{\frac{1}{2}y(y + 2)}=\frac{2}{y}$.

Step2: Evaluate the integral

We need to evaluate $\int_{\frac{1}{2}}^{1}\frac{y + 2}{\frac{1}{2}y^{2}+y}dy=\int_{\frac{1}{2}}^{1}\frac{2}{y}dy$. Using the integral formula $\int\frac{1}{y}dy=\ln|y|+C$, we have $2\int_{\frac{1}{2}}^{1}\frac{1}{y}dy$.
Applying the fundamental - theorem of calculus: $2[\ln|y|]_{\frac{1}{2}}^{1}=2(\ln(1)-\ln(\frac{1}{2}))$.
Since $\ln(1) = 0$, then $2(0-\ln(\frac{1}{2}))=2\ln(2)$.

Answer:

$2\ln(2)$