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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 \(x = -1\)
\(1 = A(-1 + 1) + \textit{B}[1 - (-1)]\)
\(1 = A(0) + \textit{B}(1 + 1)\)
\(B = \frac{1}{2}\)
\(\frac 1{(1-x)(1+x)}=\frac A{1-x}+\frac B{1+x}\)
\(= \frac{\frac 1 2}{1-x}+\frac{\frac 1 2}{1+x}\)
\(\frac 1{2(1-x)}+\frac 1{2(1+x)}\)
\(= \frac 1{(1-x)(1+x)}=\frac 1{1-x^2}=\frac 1{2(1-x)}+\frac 1{2(1+x)}\)

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Discussion (2)

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

    Reply
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  • \(\frac 1{1-x^2}\)
    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 \(x = -1\)
    \(1 = A(-1 + 1) + \textit{B}[1 - (-1)]\)
    \(1 = A(0) + \textit{B}(1 + 1)\)
    \(B = \frac{1}{2}\)
    \(\frac 1{(1-x)(1+x)}=\frac A{1-x}+\frac B{1+x}\)
    \(= \frac{\frac 1 2}{1-x}+\frac{\frac 1 2}{1+x}\)
    \(\frac 1{2(1-x)}+\frac 1{2(1+x)}\)
    \(= \frac 1{(1-x)(1+x)}=\frac 1{1-x^2}=\frac 1{2(1-x)}+\frac 1{2(1+x)}\)