Mathematics

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

##### ANSWER

$\displaystyle \frac{7}{4}$

##### SOLUTION
$\displaystyle \frac{3x+4}{(x+1)^2(x-1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$

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

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

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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