decomposition into partial fractions

could you please help me with following partial fractions ?

1) $$\frac{1}{x(x^2+1)}$$

2) $$\frac{1}{1+x^3}$$

3) $$\frac{x}{x^3-1}$$

Thank you for very much for your time

-
This sounds like a homework problem. What have you already done? –  Jebruho Nov 28 '12 at 16:28
Yes it is homework, i have already done 9 previous partial fractions but in case of these three i just dont know how to decompose. –  Radoslav Ani Smolík Nov 28 '12 at 16:32

(1) For the first one the denominator is $x \cdot (x - i) \cdot (x + i)$ so we want to find numbers $A,B,C$ such that $$\frac{1}{x(x^2+1)} = \frac{A}{x}+ \frac{B}{x-i} + \frac{C}{x+i}.$$

Multiply both sides by $x(x^2+1)$ to get $$1 = (A + B + C)x^2 + i(B - C)x + A = 1$$ reading off from this we find $A=1$, $B=C$ and $1+2B=0$ so $B = -\frac{1}{2}$.

We thus have $$\frac{1}{x(x^2+1)} = \frac{1}{x} - \frac{1}{2(x-i)} + \frac{1}{2(x+i)}$$

(3) The denominator is $(x-1)(x^2+x+1)$ so we want to find numbers $A,B,C$ such that $$\frac{x}{x^3-1} = \frac{A}{x-1} + \frac{Bx + C}{x^2+x+1}$$ multiplying both sides by $x^3-1$ gives $$x = (A + B)x^2 + (A - B + C)x + (A - C)$$ thus $A=C$, $A=-B$ and $3A=1$ so we have $$\frac{x}{x^3-1} = \frac{1}{3(x-1)} + \frac{1-x}{3(x^2+x+1)}.$$

-

The first one is standard: $x^2+1$ is an irreducible quadratic (over the reals), so you set it up as $$\frac1{x(x^2+1)}=\frac{A}x+\frac{Bx+C}{x^2+1}\;.$$ As a quick check, note that you have three unknown coefficients, exactly as you should for a denominator of degree $3$. Recombining the righthand side over its lowest common denominator, we get

$$\frac1{x(x^2+1)}=\frac{A}x+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+(Bx+C)x}{x(x^2+1)}$$

and therefore $A(x^2+1)+(Bx+C)x=1$. Multiply out and equate coefficients of like powers of $x$: $(A+B)x^2+Cx+A=1$, so

\left\{\begin{align*} A&=1&&\text{from the constant term}\\ C&=0&&\text{from the }x\text{ term}\\ A+B&=0&&\text{from the }x^2\text{ term}\;. \end{align*}\right.

Clearly $B=-1$, and the partial fraction decomposition is therefore

$$\frac1{x(x^2+1)}=\frac1x-\frac{x}{x^2+1}\;.$$

Each of the other two requires you to use a factorization identity; the identities in question are $$a^3+b^3=(a+b)(a^2-ab+b^2)$$ and $$a^3-b^3=(a-b)(a^2+ab+b^2)\;,$$ both of which are well worth memorizing. Then

$$\frac1{1+x^3}=\frac1{(1+x)(1-x+x^2)}=\frac{A}{1+x}+\frac{Bx+C}{1-x+x^2}$$

and

$$\frac{x}{x^3-1}=\frac{x}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}\;,$$

since the quadratics $x^2-x+1$ and $x^2+x+1$ are both irreducible, and you proceed as in the first problem.

-
Thank you very much. –  Radoslav Ani Smolík Nov 29 '12 at 14:53
@Radoslav: You’re welcome. –  Brian M. Scott Nov 29 '12 at 18:33