Resolve into partial fraction : $\dfrac{4}{x^2-1}$.

Resolve into partial fraction : (4)/(x²-1).

Question 23 of 48

\dfrac{4}{x^2-1}

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2020Subjective· 2mImportance★★★★★

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Factorising the denominator as $(x-1)(x+1)$ and solving gives $\dfrac{4}{x^2-1}=\dfrac{2}{x-1}-\dfrac{2}{x+1}$ .

The denominator factorises as a difference of squares:

$x^2 - 1 = (x-1)(x+1).$

Since these are two distinct (non-repeated) linear factors, write:

$\frac{4}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}.$

Multiply both sides by $(x-1)(x+1)$ :

$4 = A(x+1) + B(x-1).$

Put $x=1$ : $\;4 = A(2) + B(0) \Rightarrow 2A = 4 \Rightarrow A = 2.$

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