Basic Questions Concerning Partial Fractions

Question

Express $\displaystyle \frac x{(x^2+1)(1+x)}$ in partial fractions, I'm having a little difficulty with this. Can you tell me what the two numerators should be please.

Answer 1

Using Partial Fraction Decomposition formula,

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

as for each terms, the highest power of $x$ in the numerator will be $=$ the highest power of $x$ in the denominator $-1$

Multiplying either sides by $(x^2+1)(1+x),$

we get $$x=A(x^2+1)+(x+1)(Bx+C)\iff x=(A+B)x^2+(B+C)x+(C+A)$$

Compare the coefficients of $x,x^2$ and the constants to find $A,B,C$

Answer 2

First of all, please review the Wolfram's entry on Partial Fraction Decomposition.

$$\frac x{(x^2+1)(x+1)}=\frac {Ax + B}{x^2+1}+\frac{C}{x+1}\tag{1}$$

Note for fraction with a degree-$2$ denominator, we use $Ax + B$, which is a degree-$1$ polynomial, in the numerator. For the fraction with a degree-$1$ denominator, we use a constant $C$ (degree-$0$ polynomial) in the numerator.

Now, we need to satisfy the following equality: $$(Ax + B)(x + 1) + C(x^2 + 1) = x = 0\cdot x^2 + (1)x + 0\tag{2}$$

We can expand the left-hand side, and match of coefficients on the right-hand side, or we can carefully, strategically substitute appropriate values of $x$ to "zero out" factor(s), and/or simplify the solutions for the the coefficients $A, B, C$.

For example, when $x = -1$, substituting into equation $(2)$, we have $$(-A + B)\cdot 0 + 2C = -1 \iff C = -\frac 12.$$

When $x = 0$, substituting into equation $(2)$, and using our found value for $C$, we have $$(0A + B)(1) + C = 0 \iff B - \frac 12 = 0 \iff B = \frac 12$$

When $x = 1$, subsitituting into equation $(2)$, and using our found values for $B, C$, we have $$(A + B)(2) + 2C = 2A + 2\left(\frac 12\right) + 2\left(-\frac 12\right) = 1 \iff 2A = 1 \iff A = \frac 12.$$

Now that we've found $A, B, C$, we can complete the partial fraction decomposition $(1)$:

$$\frac x{(x^2+1)(x+1)}=\frac {Ax + B}{x^2+1}+\frac{C}{x+1} = \frac{x + 1}{2(x^2 + 1)} - \frac 1{2(x + 1)}$$