$int_{-1}^{3}\frac{x + 4}{x^{2}+2x}dx$

Explanation:

Step1: Factor the denominator

First, factor the denominator $x^{2}+2x=x(x + 2)$. Then, use partial - fraction decomposition. Let $\frac{x + 4}{x(x + 2)}=\frac{A}{x}+\frac{B}{x + 2}$. Cross - multiply: $x + 4=A(x + 2)+Bx$. Let $x=0$, then $4 = 2A$, so $A = 2$. Let $x=-2$, then $-2 + 4=-2B$, so $B=-1$. So $\frac{x + 4}{x(x + 2)}=\frac{2}{x}-\frac{1}{x + 2}$.

Step2: Integrate term - by - term

$\int_{-1}^{3}\frac{x + 4}{x^{2}+2x}dx=\int_{-1}^{3}(\frac{2}{x}-\frac{1}{x + 2})dx=\int_{-1}^{3}\frac{2}{x}dx-\int_{-1}^{3}\frac{1}{x + 2}dx$. The antiderivative of $\frac{2}{x}$ is $2\ln|x|$ and the antiderivative of $\frac{1}{x + 2}$ is $\ln|x + 2|$.

Step3: Apply the fundamental theorem of calculus

$[2\ln|x|-\ln|x + 2|]_{-1}^{3}=(2\ln3-\ln5)-(2\ln1-\ln1)$. Since $\ln1 = 0$, we have $2\ln3-\ln5=\ln3^{2}-\ln5=\ln\frac{9}{5}$.

Answer:

$\ln\frac{9}{5}$