Answer to Question #223095 in Calculus for Hayri

Question

Decompose the following into partial fractions


1) 8x3+13x / (x2+2)2


2) 4x2+5x+3 / (x+1)(x2+5)

Answer 1

$\frac{8 x^{3} + 13 x}{\left(x^{2} + 2\right)^{2}}=\frac{A x + B}{x^{2} + 2}+\frac{C x + D}{\left(x^{2} + 2\right)^{2}} = \frac{\left(x^{2} + 2\right) \left(A x + B\right) + C x + D}{\left(x^{2} + 2\right)^{2}} \\ \implies \\ 8 x^{3} + 13 x =\left(x^{2} + 2\right) \left(A x + B\right) + C x + D \\ \Leftrightarrow 8 x^{3} + 13 x=x^{3} A + x^{2} B + x \left(2 A + C\right) + 2 B + D \\ \text{Comparing the coefficients we have the following system of equation: } \\ \begin{cases} A = 8\\B = 0\\2 A + C = 13\\2 B + D = 0 \end{cases} \\ \text{Solving the system we get } A=8, B=0, C=-3, D=0 \\ \therefore \boxed{\frac{8 x^{3} + 13 x}{\left(x^{2} + 2\right)^{2}}=\frac{8 x}{x^{2} + 2}-\frac{3 x}{\left(x^{2} + 2\right)^{2}}} \\ \\ \frac{4 x^{2} + 5 x + 3}{\left(x + 1\right) \left(x^{2} + 5\right)}=\frac{A}{x + 1}+\frac{B x + C}{x^{2} + 5} = \frac{\left(x + 1\right) \left(B x + C\right) + \left(x^{2} + 5\right) A}{\left(x + 1\right) \left(x^{2} + 5\right)} \\ \Leftrightarrow 4 x^{2} + 5 x + 3=x^{2} \left(A + B\right) + x \left(B + C\right) + 5 A + C \\ \text{Comparing the coefficients we have the following system of equation: } \\ \begin{cases} A + B = 4\\B + C = 5\\5 A + C = 3 \end{cases} \\ \text{Solving these system gives: }A=\frac{1}{3}, B=\frac{11}{3}, B=\frac{4}{3} \\ \therefore \\ \boxed{\frac{4 x^{2} + 5 x + 3}{\left(x + 1\right) \left(x^{2} + 5\right)}=\frac{1}{3(x + 1)}+\frac{{11 x} + {4}}{3(x^{2} + 5)}}$ The answer is in the picture below