Resolve the following fraction into partial fractions : $\frac{2 x + 1}{(x - 1) (x^{2} + 1)}$ .

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Solution

Let, $\frac{2 x + 1}{(x - 1) (x^{2} + 1)} = \frac{A}{(x - 1)} + \frac{B x + C}{(x^{2} + 1)} . . . (1)$

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

$2 x + 1 = A (x^{2} + 1) + B (x^{2} - x) + C (x - 1) . . . . (2)$

Putting $x - 1 = 0$ i.e, $x = 1$ in equation $(2)$

$2 (1) + 1 = A (1^{2} + 1) + B (1^{2} - 1) + C (1 - 1)$

$3 = 2 A + B (0) + C (0)$

$A = \frac{3}{2}$

Now, on comparing with coefficients of $x^{2}$ in equation $(2)$

$0 = A + B$

On putting value of $A$

$0 = \frac{3}{2} + B$

$B = \frac{- 3}{2}$

On comparing with coeff. of $x$ in equation $(2)$

$2 = - B + C$

On putting value of $B$

$2 = \frac{3}{2} + C$

$C = \frac{4 - 3}{2} \Rightarrow C = \frac{1}{2}$

On putting values of $A, B$ and $C$ is equation $(1)$

$\frac{2 x + 1}{(x - 1) (x^{2} + 1)} = \frac{1}{2} [\frac{3}{x - 1} - \frac{3 x - 1}{x^{2} + 1}]$ .

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