how to work out partial fraction decomposition when given

1/x(x^2+1) and 2x-1/(x+4)(x-3)

if you could give me an idea of the formula and what changes i need to make to the formula when the denominator changes that would help and please use idiot speak because the text books are confusing me

i think i worked out the second one but i still cant do the first

is it = 9/7(x+4)+5/7(x-3)???

3 answers

let 1/x(x^2+1)
= A/x + (Bx+C)/(x^2+1)
= [A(x^2+1) + x(Bx+C)]/[x(x^2+1)]
= [Ax^2 + A + Bx^2 + Cx}/[x(x^2+1)]
= [x^2(A+B) + Cx + A]/[x(x^2+1)]

there is no x term in the original numerator, so C = 0
there is no x^2 in the original numerator, so A+B = 0,
The constant was 1, so A = 1
then B = -1, since A+B=0

so 1/x(x^2+1)
= 1/x -x/(x^2 + 1)
careful how you write the second,
it should be

9/(7(x+4))+5/(7(x-3))
What happened to the grocer who stacked all of the liquid detergent no the high shelf ?
Similar Questions
  1. how to work out partial fraction decomposition when given1/x(x^2+1) and 2x-1/(x+4)(x-3) if you could give me an idea of the
    1. answers icon 1 answer
  2. ∫ (2x+1)/(x+1) dxPlease show steps if you work it. I think it's a partial fraction and worked with decomposition. Not sure
    1. answers icon 2 answers
    1. answers icon 3 answers
    1. answers icon 2 answers
more similar questions