# How do you express ( 1+2x )/( (6x^2+1)(1-3x)) in partial fractions?

Jan 14, 2017

The answer is $= \frac{2 x}{6 {x}^{2} + 1} + \frac{1}{1 - 3 x}$

#### Explanation:

Let's do the decomposition into partial fractions

$\frac{1 + 2 x}{\left(6 {x}^{2} + 1\right) \left(1 - 3 x\right)} = \frac{A x + B}{6 {x}^{2} + 1} + \frac{C}{1 - 3 x}$

$= \frac{\left(A x + B\right) \left(1 - 3 x\right) + C \left(6 {x}^{2} + 1\right)}{\left(6 {x}^{2} + 1\right) \left(1 - 3 x\right)}$

Therefore,

$1 + 2 x = \left(A x + B\right) \left(1 - 3 x\right) + C \left(6 {x}^{2} + 1\right)$

Let $x = 0$, $\implies$, $1 = B + C$

Let $x = \frac{1}{3}$, $\implies$, $\frac{5}{3} = \frac{5}{3} C$, $\implies$, $C = 1$

So, $B = 0$

Coefficients of ${x}^{2}$, $\implies$, $0 = - 3 A + 6 C$, $\implies$, $A = 2$

Therefore,

$\frac{1 + 2 x}{\left(6 {x}^{2} + 1\right) \left(1 - 3 x\right)} = \frac{2 x}{6 {x}^{2} + 1} + \frac{1}{1 - 3 x}$

Jan 14, 2017

The Reqd. Partial Decomposition is $\frac{1}{1 - 3 x} + \frac{2 x}{6 {x}^{2} + 1}$.

#### Explanation:

Let, $f \left(x\right) = \frac{1 + 2 x}{\left(6 {x}^{2} + 1\right) \left(1 - 3 x\right)} = \frac{A}{1 - 3 x} + \frac{B x + C}{6 {x}^{2} + 1} \ldots \left(\ast\right)$,

where, consts. $A , B , C \in \mathbb{R}$.

We use Heavyside's Method to find $A , B , C$.

To find the const. $A$ corresponding to the linear factor $\left(1 - 3 x\right)$, put

$\left(1 - 3 x\right) = 0 \text{ & get } x = \frac{1}{3} ,$ and substitute this value in $f \left(x\right)$

barring $\left(1 - 3 x\right) , \text{ i.e., in } \frac{1 + 2 x}{6 {x}^{2} + 1}$.

I denote this as : $A = {\left[\frac{1 + 2 x}{6 {x}^{2} + 1}\right]}_{x = \frac{1}{3}}$

$\therefore A = \frac{1 + \frac{2}{3}}{\frac{6}{9} + 1} = \frac{\frac{5}{3}}{\frac{5}{3}} = 1$.

Then, by $\left(\ast\right) , \frac{1 + 2 x}{\left(6 {x}^{2} + 1\right) \left(1 - 3 x\right)} - \frac{1}{1 - 3 x} = \frac{B x + C}{6 {x}^{2} + 1}$

$\Rightarrow \frac{1 + 2 x - 6 {x}^{2} - 1}{\left(6 {x}^{2} + 1\right) \left(1 - 3 x\right)} = \frac{B x + C}{6 {x}^{2} + 1}$

$\Rightarrow \frac{2 x \left(1 - 3 x\right)}{\left(6 {x}^{2} + 1\right) \left(1 - 3 x\right)} = \frac{B x + C}{6 {x}^{2} + 1}$

Clearly, $B = 2 , \mathmr{and} , C = 0$.

Thus, the Reqd. Partial Decomposition of $f \left(x\right)$ is,

$\frac{1}{1 - 3 x} + \frac{2 x}{6 {x}^{2} + 1}$, in accordance with that of Respected