$\int \large\frac{3x}{(x+1)(x+2)}$$dx$
first find x value in denominator,
$x=-1,x=-2$
Put x value in denominator and numerator and do put value of x, where infinite condition occured so,
$\int \large\frac{3(-2)}{(-2+1)(x+2)}$$dx+\int \large\frac{3(-1)}{(x-1)(-1+2)}$
$\qquad= \int \large\frac{6}{(x+1)}$$dx-\int \large\frac{3}{x+1}$$dx$
$6 \log (x+2)-3 \log (x+1)+c$
Hence a is the correct answer.