Question

If $\frac{3x+4}{(x+1{)}^2(x-1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}$, then $A$=

• A. $\frac{1}{2}$
• B. $\frac{15}{4}$
• C. $\frac{7}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{7}{4}$

$\frac{3x+4}{(x+1{)}^2(x-1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1{)}^2}$
$\frac{3x+4}{(x+1{)}^2(x-1)}=\frac{A(x+1{)}^2+B(x^2-1)+C(x-1)}{(x+1{)}^2(x-1)}$
$\Rightarrow 3x+4=A(x+1{)}^2+B(x^2-1)+C(x-1)$
When $x=1,7=4A$
$\Rightarrow A=\frac{7}{4}$