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{3 x + 2}{x (x - 2) (x + 1)}

into partial fractions.

Accepted Answer

Given $\frac{3 x + 2}{x (x - 2) (x + 1)}$

It is of the form $\frac{f (x)}{g (x)}$ where $g (x)$ contains non-repeated linear factors. That is $g (x) = (x - α_{1}) (x - α_{2}) (x - α_{3})$ .

$\Rightarrow \frac{f (x)}{g (x)} = \frac{A}{(x - α_{1})} + \frac{B}{(x - α_{2})} + \frac{C}{(x - α_{3})}$

Now, $\frac{3 x + 2}{x (x - 2) (x + 1)} = \frac{A}{x} + \frac{B}{(x - 2)} + \frac{C}{(x + 1)} . . . . . . . (i)$

$\Rightarrow 3 x + 2 = (x - 2) (x + 1) A + x (x + 1) B + x (x - 2) C$

$\Rightarrow 3 x + 2 = (x^{2} - x - 2) A + (x^{2} + x) B + (x^{2} - 2 x) C$

$\Rightarrow 3 x + 2 = (A + B + C) x^{2} + (- A + B - 2 C) x - 2 A . . . . . . (ii)$

Comparing the constants,

$\Rightarrow - 2 A = 2 \Rightarrow A = - 1$

Comparing the coefficients of $x^{2}$ and $x$ of $(ii)$ equation,

$A + B + C = 0 \Rightarrow B + C = 1 . . . . . . (iii)$

And, $- A + B - 2 C = 3$

$\Rightarrow 1 + (1 - C) - 2 C = 3 \Rightarrow - 3 C = 1 \Rightarrow C = - \frac{1}{3}$

From $(iii)$ equation, $B - \frac{1}{3} = 1$

$\Rightarrow B = \frac{4}{3}$

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

Now, $\frac{3 x + 2}{x (x - 2) (x + 1)} = \frac{- 1}{x} + \frac{4}{3 (x - 2)} - \frac{1}{3 (x + 1)}$

Resolve $\frac{3 x + 2}{x (x - 2) (x + 1)}$ into partial fractions.

Important Questions on Partial Fractions

Learn and Prepare for any exam you want !

Resolve 3x+2xx-2x+1 into partial fractions.

Important Questions on Partial Fractions

Learn and Prepare for any exam you want !

Resolve $\frac{3 x + 2}{x (x - 2) (x + 1)}$ into partial fractions.