Partial Fraction Decomposition Calculator

Your input: perform the partial fraction decomposition of $$$\frac{2 x}{\left(x + 1\right)^{2} \left(x^{2} + 1\right)}$$$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

The denominators are equal, so we require the equality of the numerators:

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

Expand the right-hand side:

$$2 x=x^{3} A + x^{3} C + x^{2} A + x^{2} B + 2 x^{2} C + x^{2} D + x A + x C + 2 x D + A + B + D$$

Collect up the like terms:

$$2 x=x^{3} \left(A + C\right) + x^{2} \left(A + B + 2 C + D\right) + x \left(A + C + 2 D\right) + A + B + D$$

The coefficients near the like terms should be equal, so the following system is obtained:

$$\begin{cases} A + C = 0\\A + B + 2 C + D = 0\\A + C + 2 D = 2\\A + B + D = 0 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=0$$$, $$$B=-1$$$, $$$C=0$$$, $$$D=1$$$

Therefore,

$$\frac{2 x}{\left(x + 1\right)^{2} \left(x^{2} + 1\right)}=\frac{0}{x + 1}+\frac{-1}{\left(x + 1\right)^{2}}+\frac{1}{x^{2} + 1}=\frac{-1}{\left(x + 1\right)^{2}}+\frac{1}{x^{2} + 1}$$

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