\section*{Exercise 3 - 111013}

Let \(x\) be a variable that can take all real values except 1 and -1.

Provide a way to express the term \(\frac{x}{x^{2}-1}\) as a sum of two fractions such that the variable \(x\) appears only in the denominators of these two fractions and only in the first power!
}

It is natural to choose the denominators \(x+1\) and \(x-1\) for the sought quotients, because their least common denominator is equal to the denominator of the given term. Therefore, we choose the ansatz

\[
\frac{x}{x^{2}-1}=\frac{a}{x+1}+\frac{b}{x-1}
\]

This implies

\[
\frac{x}{x^{2}-1}=\frac{a x-a+b x+b}{x^{2}-1}
\]

Thus, after multiplying by \(x^{2}-1\),

\[
x=a x-a+b x+b=x(a+b)-a+b
\]

For \(x=0\), it follows that \(-a+b=0 \quad\) (2) must hold. From (1) and (2), it further follows that \(x=x(a+b)\) and thus, since this must hold for at least one \(x \neq 0\): \(a+b=1 \quad\) (3).

The system of equations (2), (3) has only the solution \(a=b=\frac{1}{2}\). The verification confirms the found decomposition

\[
\frac{1}{2(x+1)}+\frac{1}{2(x-1)}=\frac{x-1+x+1}{2(x+1)(x-1)}=\frac{1}{x^{2}-1}
\]