Question:
$\frac{3 x-1}{(x+2)^{2}}$
Solution:
Let $\frac{3 x-1}{(x+2)^{2}}=\frac{A}{(x+2)}+\frac{B}{(x+2)^{2}}$
$\Rightarrow 3 x-1=A(x+2)+B$
Equating the coefficient of x and constant term, we obtain
A = 3
2A + B = −1 ⇒ B = −7
$\therefore \frac{3 x-1}{(x+2)^{2}}=\frac{3}{(x+2)}-\frac{7}{(x+2)^{2}}$
$\Rightarrow \int \frac{3 x-1}{(x+2)^{2}} d x=3 \int \frac{1}{(x+2)} d x-7 \int \frac{x}{(x+2)^{2}} d x$
$=3 \log |x+2|-7\left(\frac{-1}{(x+2)}\right)+\mathrm{C}$
$=3 \log |x+2|+\frac{7}{(x+2)}+\mathrm{C}$
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