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

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Answer 1 · 2017-01-14T16:16:52

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

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 · 2017-01-14T17:10:15

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

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.