evaluate the integral. (remember to use absolute values where appropriate. use c for the constant of…

Explanation:

Step1: Perform partial - fraction decomposition

Let $\frac{5x + 1}{(2x + 1)(x - 1)}=\frac{A}{2x+1}+\frac{B}{x - 1}$. Then $5x + 1=A(x - 1)+B(2x + 1)$. Set $x = 1$, we get $5\times1+1=A(1 - 1)+B(2\times1 + 1)$, so $6 = 3B$, and $B = 2$. Set $x=-\frac{1}{2}$, we get $5\times(-\frac{1}{2})+1=A(-\frac{1}{2}-1)+B(2\times(-\frac{1}{2})+1)$. $-\frac{5}{2}+1=A(-\frac{3}{2})+B(0)$, $-\frac{3}{2}=-\frac{3}{2}A$, so $A = 1$. So $\frac{5x + 1}{(2x + 1)(x - 1)}=\frac{1}{2x+1}+\frac{2}{x - 1}$.

Step2: Integrate term - by - term

$\int\frac{5x + 1}{(2x + 1)(x - 1)}dx=\int\frac{1}{2x+1}dx+\int\frac{2}{x - 1}dx$. For $\int\frac{1}{2x+1}dx$, let $u = 2x+1$, then $du=2dx$, and $\int\frac{1}{2x+1}dx=\frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}\ln|u|=\frac{1}{2}\ln|2x + 1|$. For $\int\frac{2}{x - 1}dx=2\int\frac{1}{x - 1}dx=2\ln|x - 1|$. So $\int\frac{5x + 1}{(2x + 1)(x - 1)}dx=\frac{1}{2}\ln|2x + 1|+2\ln|x - 1|+C$.

Answer:

$\frac{1}{2}\ln|2x + 1|+2\ln|x - 1|+C$