Question

$\frac{3x^2+5}{(x^2+1{)}^2}=\frac{A}{(x^2+1)}+\frac{B}{(x^2+1{)}^2}\Rightarrow (A,B)=$

• A. $(2,3)$
• B. $(3,2)$
• C. $(2,-3)$
• D. $(-2,3)$

Answer 1

The correct option is B $(3,2)$

solving through substitution
$x^2+1=k\Rightarrow x^2=k-1$
substituting in the eqn. we get
$\frac{3(k-1)+5}{k^2}=\frac{3k+2}{k^2}=\frac{3}{k}+\frac{2}{k^2}$
$=\frac{3}{(x^2+1)}+\frac{2}{(x^2+1{)}^2}$ since $k=x^2+1$
So, $A=3;B=2$
$(A,B)=(3,2)$