8 x 3 + 13 x ( x 2 + 2 ) 2 = A x + B x 2 + 2 + C x + D ( x 2 + 2 ) 2 = ( x 2 + 2 ) ( A x + B ) + C x + D ( x 2 + 2 ) 2 ⟹ 8 x 3 + 13 x = ( x 2 + 2 ) ( A x + B ) + C x + D ⇔ 8 x 3 + 13 x = x 3 A + x 2 B + x ( 2 A + C ) + 2 B + D Comparing the coefficients we have the following system of equation: { A = 8 B = 0 2 A + C = 13 2 B + D = 0 Solving the system we get A = 8 , B = 0 , C = − 3 , D = 0 ∴ 8 x 3 + 13 x ( x 2 + 2 ) 2 = 8 x x 2 + 2 − 3 x ( x 2 + 2 ) 2 4 x 2 + 5 x + 3 ( x + 1 ) ( x 2 + 5 ) = A x + 1 + B x + C x 2 + 5 = ( x + 1 ) ( B x + C ) + ( x 2 + 5 ) A ( x + 1 ) ( x 2 + 5 ) ⇔ 4 x 2 + 5 x + 3 = x 2 ( A + B ) + x ( B + C ) + 5 A + C Comparing the coefficients we have the following system of equation: { A + B = 4 B + C = 5 5 A + C = 3 Solving these system gives: A = 1 3 , B = 11 3 , B = 4 3 ∴ 4 x 2 + 5 x + 3 ( x + 1 ) ( x 2 + 5 ) = 1 3 ( x + 1 ) + 11 x + 4 3 ( x 2 + 5 ) \frac{8 x^{3} + 13 x}{\left(x^{2} + 2\right)^{2}}=\frac{A x + B}{x^{2} + 2}+\frac{C x + D}{\left(x^{2} + 2\right)^{2}} = \frac{\left(x^{2} + 2\right) \left(A x + B\right) + C x + D}{\left(x^{2} + 2\right)^{2}} \\
\implies \\
8 x^{3} + 13 x =\left(x^{2} + 2\right) \left(A x + B\right) + C x + D \\
\Leftrightarrow 8 x^{3} + 13 x=x^{3} A + x^{2} B + x \left(2 A + C\right) + 2 B + D \\
\text{Comparing the coefficients we have the following system of equation: } \\
\begin{cases} A = 8\\B = 0\\2 A + C = 13\\2 B + D = 0 \end{cases} \\
\text{Solving the system we get } A=8, B=0, C=-3, D=0 \\
\therefore \boxed{\frac{8 x^{3} + 13 x}{\left(x^{2} + 2\right)^{2}}=\frac{8 x}{x^{2} + 2}-\frac{3 x}{\left(x^{2} + 2\right)^{2}}} \\ \\
\frac{4 x^{2} + 5 x + 3}{\left(x + 1\right) \left(x^{2} + 5\right)}=\frac{A}{x + 1}+\frac{B x + C}{x^{2} + 5} = \frac{\left(x + 1\right) \left(B x + C\right) + \left(x^{2} + 5\right) A}{\left(x + 1\right) \left(x^{2} + 5\right)} \\
\Leftrightarrow 4 x^{2} + 5 x + 3=x^{2} \left(A + B\right) + x \left(B + C\right) + 5 A + C \\
\text{Comparing the coefficients we have the following system of equation: } \\
\begin{cases} A + B = 4\\B + C = 5\\5 A + C = 3 \end{cases} \\
\text{Solving these system gives: }A=\frac{1}{3}, B=\frac{11}{3}, B=\frac{4}{3} \\
\therefore \\
\boxed{\frac{4 x^{2} + 5 x + 3}{\left(x + 1\right) \left(x^{2} + 5\right)}=\frac{1}{3(x + 1)}+\frac{{11 x} + {4}}{3(x^{2} + 5)}} ( x 2 + 2 ) 2 8 x 3 + 13 x = x 2 + 2 A x + B + ( x 2 + 2 ) 2 C x + D = ( x 2 + 2 ) 2 ( x 2 + 2 ) ( A x + B ) + C x + D ⟹ 8 x 3 + 13 x = ( x 2 + 2 ) ( A x + B ) + C x + D ⇔ 8 x 3 + 13 x = x 3 A + x 2 B + x ( 2 A + C ) + 2 B + D Comparing the coefficients we have the following system of equation: ⎩ ⎨ ⎧ A = 8 B = 0 2 A + C = 13 2 B + D = 0 Solving the system we get A = 8 , B = 0 , C = − 3 , D = 0 ∴ ( x 2 + 2 ) 2 8 x 3 + 13 x = x 2 + 2 8 x − ( x 2 + 2 ) 2 3 x ( x + 1 ) ( x 2 + 5 ) 4 x 2 + 5 x + 3 = x + 1 A + x 2 + 5 B x + C = ( x + 1 ) ( x 2 + 5 ) ( x + 1 ) ( B x + C ) + ( x 2 + 5 ) A ⇔ 4 x 2 + 5 x + 3 = x 2 ( A + B ) + x ( B + C ) + 5 A + C Comparing the coefficients we have the following system of equation: ⎩ ⎨ ⎧ A + B = 4 B + C = 5 5 A + C = 3 Solving these system gives: A = 3 1 , B = 3 11 , B = 3 4 ∴ ( x + 1 ) ( x 2 + 5 ) 4 x 2 + 5 x + 3 = 3 ( x + 1 ) 1 + 3 ( x 2 + 5 ) 11 x + 4
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