Answer
$\displaystyle \frac{1}{2}\ln|2x+1|+2\ln|x-1| + C$
Work Step by Step
Partial Fraction Decomposition: $\displaystyle \quad\frac{px+q}{(x-a)(x-b)} = \frac{A}{x-a}+\frac{B}{x-b}$
$\displaystyle \int\frac{5x+1}{(2x+1)(x-1)}dx$
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Partial Fraction Decomposition:
$5x+1=A(2x+1)+B(x-1)$
Let $x = \displaystyle -\frac{1}{2}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$Let $x=1$
$\displaystyle 5(-\frac{3}{2}) + 1=B(-\frac{1}{2}-1)\quad\quad\quad\quad\quad\quad\quad 5(1)+1=A(2(1)+1)$
$\displaystyle -\frac{3}{2}=-\frac{3}{2}B\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad 6 = 3A$
$1=B\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad2=A$
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$\displaystyle \int\frac{5x+1}{(2x+1)(x-1)}dx = \int\frac{1}{2x+1}+\frac{2}{x-1}dx = \int\frac{2}{x-1}dx+\int\frac{1}{2x+1}dx$
$\displaystyle 2\ln|x-1|+\int\frac{1}{2x+1}dx$, Let $u = 2x+1\quad \rightarrow \quad du=2dx$
$\displaystyle 2\ln|x-1|+\frac{1}{2}\int\frac{1}{u}du$
$\displaystyle 2\ln|x-1| + \frac{1}{2}\ln|u|+ C$
$\displaystyle 2\ln|x-1| + \frac{1}{2}\ln|2x+1|+ C$
