Question

solve 1/(x+1)(x^2-1) into partial​

Answer 1

Answer:

                 $\displaystyle-\frac1{4(x+1)}-\frac1{2(x+1)^2}+\frac1{4(x-1)}$

Step-by-step explanation:

$\displaystyle\frac1{(x+1)(x^2-1)}=\frac1{(x+1)^2(x-1)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x-1}\\\\\Rightarrow 1 = A(x^2-1)+B(x-1)+C(x+1)^2$

Putting x = 1 gives  1 = 0 + 0 + 4C  ⇒  C = 1/4.

Putting x = -1 gives  1 = 0 - 2B + 0  ⇒  B = -1/2.

Putting x = 0 gives  1 = -A - B + C  ⇒  A = C - B - 1  =  1/4 + 1/2 - 1  =  -1/4.

Therefore

$\displaystyle\frac1{(x+1)(x^2-1)}=-\frac{\frac14}{x+1}-\frac{\frac12}{(x+1)^2}+\frac{\frac14}{x-1}\\\\=-\frac1{4(x+1)}-\frac1{2(x+1)^2}+\frac1{4(x-1)}$