Question

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

Answer 1

The answer is `=(2x)/(6x^2+1)+1/(1-3x)`

Explanation:

Let's do the decomposition into partial fractions

`(1+2x)/((6x^2+1)(1-3x))=(Ax+B)/(6x^2+1)+C/(1-3x)`

`=((Ax+B)(1-3x)+C(6x^2+1))/((6x^2+1)(1-3x))`

Therefore,

`1+2x=(Ax+B)(1-3x)+C(6x^2+1)`

Let `x=0`, `=>`, `1=B+C`

Let `x=1/3`, `=>`, `5/3=5/3C`, `=>`, `C=1`

So, `B=0`

Coefficients of `x^2`, `=>`, `0=-3A+6C`, `=>`, `A=2`

Therefore,

`(1+2x)/((6x^2+1)(1-3x))=(2x)/(6x^2+1)+1/(1-3x)`

Answer 2

The Reqd. Partial Decomposition is `1/(1-3x)+(2x)/(6x^2+1)`.

Explanation:

Let, `f(x)=(1+2x)/{(6x^2+1)(1-3x)}=A/(1-3x)+(Bx+C)/(6x^2+1)...(ast)`,

where, consts. `A,B,C in RR`.

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

To find the const. `A` corresponding to the linear factor `(1-3x)`, put

`(1-3x)=0" & get "x=1/3,` and substitute this value in `f(x)`

barring `(1-3x)," i.e., in "(1+2x)/(6x^2+1)`.

I denote this as : `A=[(1+2x)/(6x^2+1)]_(x=1/3)`

`:. A=(1+2/3)/(6/9+1)=(5/3)/(5/3)=1`.

Then, by `(ast), (1+2x)/{(6x^2+1)(1-3x)}-1/(1-3x)=(Bx+C)/(6x^2+1)`

`rArr (1+2x-6x^2-1)/{(6x^2+1)(1-3x)}=(Bx+C)/(6x^2+1)`

`rArr (2x(1-3x))/{(6x^2+1)(1-3x)}=(Bx+C)/(6x^2+1)`

Clearly, `B=2, and, C=0`.

Thus, the Reqd. Partial Decomposition of `f(x)` is,

`1/(1-3x)+(2x)/(6x^2+1)`, in accordance with that of Respected

Narad T.