Example 1. Decompose into partial fractions: $\frac{1}{(x+1)\left(x^{2}+1\right)}$.
## Solution. We have

$$
\frac{1}{(x+1)\left(x^{2}+1\right)}=\frac{A}{x+1}+\frac{B x+C}{x^{2}+1}=\frac{A\left(x^{2}+1\right)+(B x+C)(x+1)}{(x+1)\left(x^{2}+1\right)}
$$

Apply Theorem 3. We arrive at the equality $1=A x^{2}+A+B x^{2}+B x+$ $+C x+C$ (the denominators of the equal fractions are the same). Now apply Theorem 4 - equate the coefficients of the same powers of polynomials on the right and left sides of the equality sign. On the right side, we have a polynomial of the second degree. We assume that on the left side, we also have a polynomial of the second degree with coefficients of positive powers of the variable $x$ equal to zero. The corresponding equalities are written in the form

$$
\left.\begin{array}{l}
x^{2}: A+B=0 \\
x^{1}: B+C=0, \\
x^{0}: A+C=1
\end{array}\right\}+\Rightarrow 2(A+B+C)=1 \Rightarrow\left\{\begin{array}{cl}
2 C=1, & C=\frac{1}{2} \\
A=\frac{1}{2}, & B=-\frac{1}{2}
\end{array}\right.
$$

$$
\text { Answer. } \frac{1}{(x+1)\left(x^{2}+1\right)}=\frac{1}{2(x+1)}+\frac{-x+1}{x^{2}+1} .
$$