Answer
$$\int^1_{1/2}\frac{y+4}{y^2+y}dy=3\ln3-2\ln2=\ln \frac{27}{4}$$
Work Step by Step
$$A=\int^1_{1/2}\frac{y+4}{y^2+y}dy$$
1) Express the integrand as a sum of partial fractions:
$$\frac{y+4}{y^2+y}=\frac{y+4}{(y+1)y}=\frac{A}{y+1}+\frac{B}{y}$$
Clear fractions:
$$y+4=Ay+By+B$$ $$y+4=(A+B)y+B$$
Equating coefficients of corresponding powers of $y$, we get
- $A+B=1$
- $B=4$
That means $A=-3$.
Therefore, $$\frac{y+4}{y^2+y}=\frac{4}{y}-\frac{3}{y+1}$$
2) Evaluate the integral: $$A=4\int^1_{1/2}\frac{dy}{y}-3\int^1_{1/2}\frac{dy}{y+1}$$ $$A=4\ln|y|\Big]^1_{1/2}-3\ln|y+1|\Big]^1_{1/2}$$ $$A=4(\ln1-\ln\frac{1}{2})-3(\ln2-\ln\frac{3}{2})$$ $$A=4(\ln1-\ln1+\ln2)-3(\ln2-\ln3+\ln2)$$ $$A=4\ln2-6\ln2+3\ln3$$ $$A=3\ln3-2\ln2=\ln \frac{27}{4}$$
