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We need to decompose the integrand into partial fractions. Let $$ \frac{2x+1}{(x+1)^2(x-1)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1} $$ Multiplying both sides by $(x+1)^2(x-1)$, we get $$ 2x+1 = A(x+1)(x-1) + B(x-1) + C(x+1)^2 $$ $$ 2x+1 = A(x^2-1) + B(x-1) + C(x^2+2x+1) $$ $$ 2x+1 = Ax^2 - A + Bx - B + Cx^2 + 2Cx + C $$ $$ 2x+1 = (A+C)x^2 + (B+2C)x + (-A-B+C) $$
Comparing coefficients, we have the following system of equations: $$ A+C = 0 $$ $$ B+2C = 2 $$ $$ -A-B+C = 1 $$ From the first equation, $A = -C$. Substituting this into the third equation: $$ -(-C) - B + C = 1 $$ $$ 2C - B = 1 $$ Adding this to the second equation $B+2C=2$, we get: $$ 4C = 3 $$ $$ C = \frac{3}{4} $$ Then, $A = -C = -\frac{3}{4}$. Substituting $C = \frac{3}{4}$ into $B+2C=2$, we get: $$ B + 2(\frac{3}{4}) = 2 $$ $$ B + \frac{3}{2} = 2 $$ $$ B = 2 - \frac{3}{2} = \frac{1}{2} $$ So, $A = -\frac{3}{4}$, $B = \frac{1}{2}$, and $C = \frac{3}{4}$.
Now we can rewrite the integral as: $$ \int \frac{2x+1}{(x+1)^2(x-1)} dx = \int \left( \frac{-\frac{3}{4}}{x+1} + \frac{\frac{1}{2}}{(x+1)^2} + \frac{\frac{3}{4}}{x-1} \right) dx $$ $$ = -\frac{3}{4} \int \frac{1}{x+1} dx + \frac{1}{2} \int \frac{1}{(x+1)^2} dx + \frac{3}{4} \int \frac{1}{x-1} dx $$ $$ = -\frac{3}{4} \ln|x+1| + \frac{1}{2} \int (x+1)^{-2} dx + \frac{3}{4} \ln|x-1| $$ $$ = -\frac{3}{4} \ln|x+1| + \frac{1}{2} \frac{(x+1)^{-1}}{-1} + \frac{3}{4} \ln|x-1| + C $$ $$ = -\frac{3}{4} \ln|x+1| - \frac{1}{2(x+1)} + \frac{3}{4} \ln|x-1| + C $$ $$ = \frac{3}{4} \ln\left|\frac{x-1}{x+1}\right| - \frac{1}{2(x+1)} + C $$
Final Answer: $\frac{3}{4} \ln\left|\frac{x-1}{x+1}\right| - \frac{1}{2(x+1)} + C$
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