Question
Integrate the rational function $\frac{3 x-1}{(x+2)^{2}}$

Answer

Let $\frac{3 x-1}{(x+2)^{2}}=\frac{A}{(x+2)}+\frac{B}{(x+2)^{2}}$ 
$\Rightarrow$ 3x -1 = A(x + 2) + B
Equating the coefficients of x and constant term, we get,
A = 3
2A + B = -1
B = -7
Thus,
$\frac{3 x-1}{(x+2)^{2}}=\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}$ 
$\Rightarrow$$\int \frac{3 x-1}{(x+2)^{2}}=\int\left\{\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}\right\} d x$ 
= $3 \log |x+2|-7\left(\frac{-1}{(x+2)}\right)+C$ 
= $3 \log |x+2|+\frac{7}{(x+2)}+C$

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