Question

Integrate the rational function: $$ \int \frac{3x - 1}{(x + 2)^{2}} \mathrm{d}x. $$

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Answer

To solve the integral $$ \int \frac{3x - 1}{(x + 2)^{2}} \mathrm{d}x $$ we start by expressing the integrand as a sum of simpler fractions. We set: $$ \frac{3x - 1}{(x + 2)^2} = \frac{A}{x + 2} + \frac{B}{(x + 2)^2} $$ This leads to: $$ 3x - 1 = A(x + 2) + B $$ Expanding and equating coefficients, we have: $$ A(x + 2) + B = Ax + 2A + B = 3x - 1 $$ From this equation, equating the coefficients of $x$ and the constant terms respectively provides: $$ A = 3 \quad \text{and} \quad 2A + B = -1 $$ Substituting $A = 3$ into the second equation gives: $$ 2(3) + B = -1 \Rightarrow 6 + B = -1 \Rightarrow B = -7 $$ Thus, the fraction decomposes into: $$ \frac{3x - 1}{(x + 2)^2} = \frac{3}{x + 2} - \frac{7}{(x + 2)^2} $$ Now, integrating each term individually: $$ \int \frac{3}{x + 2} , dx = 3 \int \frac{1}{x + 2} , dx = 3 \log|x + 2| + C_1 $$ and $$ \int \frac{-7}{(x + 2)^2} , dx = -7 \left( -\frac{1}{x + 2} \right) + C_2 = \frac{7}{x + 2} + C_2 $$ Combining both results: $$ \int \frac{3x - 1}{(x + 2)^{2}} \mathrm{d}x = 3 \log|x + 2| + \frac{7}{x + 2} + C $$ where $C$ combines the constants $C_1$ and $C_2$. Thus, the final answer is: $$ \boxed{3 \log|x + 2| + \frac{7}{x + 2} + C} $$


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