Name: Method of Partial Fractions
Uploaded: 2019-07-19
Duration: 5 min 36 s
Description: Watch this video to understand how proper rational functions are integrated when the denominator consists of a linear or a quadratic expression.

Question

Resolve

\frac{1}{(x - 1) (x^{2} + 1)}

into partial fractions.

Accepted Answer

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

For a proper fraction $\frac{f (x)}{g (x)}$ where $g (x)$ contains irreducible factors. That is $g (x) = (a x^{2} + b x + c) (x - α_{1})$ .

$\Rightarrow \frac{f (x)}{g (x)} = \frac{A x + B}{(a x^{2} + b x + c)} + \frac{C}{(x - α_{1})}$

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

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

Put $x = 1$ in equation $(ii)$ ,

$1 = (1 + 1) A + 0 \Rightarrow A = \frac{1}{2}$

Put $x = 0$ in equation $(ii)$ ,

$1 = (0 + 1) A + (0 - 1) (0 + C) \Rightarrow 1 = \frac{1}{2} - C \Rightarrow C = \frac{- 1}{2}$

Put $x = - 1$ in equation $(ii)$ ,

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

Substituting values of $A, B, C$ in equation $(i)$ ,

Now, $\frac{1}{(x - 1) (x^{2} + 1)} = \frac{1}{2 (x - 1)} + \frac{(\frac{- 1}{2}) x + (- \frac{1}{2})}{(x^{2} + 1)}$

$\Rightarrow \frac{1}{(x - 1) (x^{2} + 1)} = \frac{1}{2 (x - 1)} - \frac{(x + 1)}{2 (x^{2} + 1)}$

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