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Resolve $$\frac 1{(1-x^2)}$$ into partial fractions

Question

Resolve $$\frac 1{(1-x^2)}$$ into partial fractions

Options

A) $$\frac 1{2(1+x)}-\frac 1{2(1-x)}$$

B) $$\frac 1{2(1+x)}+\frac 1{2(x-1)}$$

C) $$\frac 1{2(x+1)}+\frac 1{2(1-x)}$$

D) $$\frac 1{2(1-x^2)}$$

The correct answer is C.

Explanation:

$$\frac 1{1-x^2}$$
But $$1 - x^2=(1 - x)(1+x)$$
$$\frac 1{1-x^2}=\frac 1{(1-x)(1+x)}$$
A linear factor of the form ax + b always gives a partial fraction of $$\frac A{\mathit{ax + b}}$$
$$\frac 1{(1-x)(1+x)}=\frac A{1-x}+\frac B{1+x}$$
$$\frac 1{(1-x)(1+x)}$$
=$$\frac{A\left(1+x\right)+B(1-x)}{(1-x)(1+x)}$$
$$1 = A(1 + x) + \textit{B}(1 - x)$$
Let $$x = 1$$
$$1 = \textit{A(}1 + 1) + \textit{B}(1 - 1)$$
$$1 = 2\textit{A} + \textit{B}(0)$$
$$1 = 2A$$
$$A = \frac{1}{2}$$
Let $$x = -1$$
$$1 = A(-1 + 1) + \textit{B}[1 - (-1)]$$
$$1 = A(0) + \textit{B}(1 + 1)$$
$$B = \frac{1}{2}$$
$$\frac 1{(1-x)(1+x)}=\frac A{1-x}+\frac B{1+x}$$
$$= \frac{\frac 1 2}{1-x}+\frac{\frac 1 2}{1+x}$$
$$\frac 1{2(1-x)}+\frac 1{2(1+x)}$$
$$= \frac 1{(1-x)(1+x)}=\frac 1{1-x^2}=\frac 1{2(1-x)}+\frac 1{2(1+x)}$$

Discussion (2)

• B is the correct option not C
You show the workings of B

Reply
• $$\frac 1{1-x^2}$$
But $$1 - x^2=(1 - x)(1+x)$$
$$\frac 1{1-x^2}=\frac 1{(1-x)(1+x)}$$
A linear factor of the form ax + b always gives a partial fraction of $$\frac A{\mathit{ax + b}}$$
$$\frac 1{(1-x)(1+x)}=\frac A{1-x}+\frac B{1+x}$$
$$\frac 1{(1-x)(1+x)}$$
=$$\frac{A\left(1+x\right)+B(1-x)}{(1-x)(1+x)}$$
$$1 = A(1 + x) + \textit{B}(1 - x)$$
Let $$x = 1$$
$$1 = \textit{A(}1 + 1) + \textit{B}(1 - 1)$$
$$1 = 2\textit{A} + \textit{B}(0)$$
$$1 = 2A$$
$$A = \frac{1}{2}$$
Let $$x = -1$$
$$1 = A(-1 + 1) + \textit{B}[1 - (-1)]$$
$$1 = A(0) + \textit{B}(1 + 1)$$
$$B = \frac{1}{2}$$
$$\frac 1{(1-x)(1+x)}=\frac A{1-x}+\frac B{1+x}$$
$$= \frac{\frac 1 2}{1-x}+\frac{\frac 1 2}{1+x}$$
$$\frac 1{2(1-x)}+\frac 1{2(1+x)}$$
$$= \frac 1{(1-x)(1+x)}=\frac 1{1-x^2}=\frac 1{2(1-x)}+\frac 1{2(1+x)}$$