91Ó°ÊÓ

We have a rational function of the form: $$\frac{20x}{(x-1)^{2}(x^2 + 1)}$$ The components of the fraction we will focus on are the polynomial numerator (20x) and the decomposed denominator (\((x-1)^{2}\) and \((x^2 + 1)\)).

In order to decompose the given rational function into partial fractions, we need to write it in the following form: $$\frac{20x}{(x-1)^{2}(x^2 + 1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1}$$ Where A, B, C, and D are constants that we need to determine.

To eliminate all denominators and find the constants, we can multiply both sides of the equation by the denominator of the original fraction. This gives us: $$20x = A(x-1)(x^2+1) + B(x^2+1) + (Cx+D)(x-1)^2$$

To find the values of the constants, we can plug in some suitable x-values to create a system of equations, or we can match coefficients of the terms in the equation obtained in step 3: Matching the coefficients of the terms, we get: $$A + B = 0$$ (for x^2) $$-A + C = 20$$ (for x) $$-A-D = 0$$ (for the constant term) We have 4 unknowns but only 3 equations. However, we can take another suitable value of x to create the fourth equation. Let's substitute x = 1, then the equation from step 3 becomes: $$20 = (C + D)$$ Now we have a system of 4 linear equations with 4 unknowns: 1. A + B = 0 2. -A + C = 20 3. -A - D = 0 4. C + D = 20 Solving this system of equations (using substitution, elimination, or matrix methods), we obtain the values of the constants: A = 10, B = -10, C = 20, and D = 0.

Now that we have the values of the constants A, B, C, and D, we can write the partial fraction decomposition of the given rational function as: $$\frac{20x}{(x-1)^{2}(x^2 + 1)} = \frac{10}{x-1} - \frac{10}{(x-1)^2} + \frac{20x}{x^2+1}$$ This is the partial fraction decomposition of the given rational function.