If $∫\frac{2x}{x^2+3x+2}dx=m log |x+2|+n log |x+1|+C, $ then the values of m and n are :

$m=4, n=-2$

The correct answer is Option (1) → $m=4, n=-2$

$∫\frac{2x}{x^2+3x+2}dx=∫\frac{2x+3}{x^2+3x+2}-\frac{3}{x^2+3x+2}dx$

$⇒\log|x^2+3x+2|-3∫\frac{1}{(x+2)(x+1)}dx$

$⇒\log|x^2+3x+2|-3∫\frac{1}{1+x}-\frac{1}{x+2}dx$

as $\frac{1}{(x+2)(x+1)}=\frac{(x+2)-(x+1)}{(x+2)(x+1)}=\frac{1}{1+x}-\frac{1}{x+2}$

$\log(x+2)(x+1)-3\log(1+x)+3\log(2+x)+C$

$=\log(x+2)+3\log(x+2)+\log(x+1)-3\log(x+1)+C$

$4\log(x+2)-2\log(x+1)+C$

$m=4,n=-2$