Question

Evaluate
$\int \frac{3x+1}{(x-2{)}^2(x+2)}dx$

Answer 1

Let $\frac{3x+1}{(x-2{)}^2(x+2)}=\frac{A}{x-2}+\frac{B}{(x-2{)}^2}+\frac{C}{x+2}$

$⟹3x+1=A(x-2)(x+2)+B(x+2)+C(x-2{)}^2$

$⟹3x+1=A(x^2-4)+B(x+2)+C(x^2-4x+4)$

$⟹A+C=0⟹A=-C$

$⟹B-4C=3,-4A+2B+4C=1⟹2B+8C=1$

Solving these we get $B=\frac{7}{4},C=-\frac{5}{16},A=\frac{5}{16}$

$\int \frac{3x+1}{(x-2{)}^2(x+2)}dx=\frac{5}{16}\int \frac{dx}{x-2}+\frac{7}{4}\int \frac{dx}{(x-2{)}^2}-\frac{5}{16}\int \frac{dx}{x+2}=\frac{5}{16}\ln |x-2|-\frac{7}{4(x-2)}-\frac{5}{16}\ln |x+2|$