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
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