Partial fraction decomposition of a complicated rational function

Question

Find the partial fraction decomposition of the rational function $\displaystyle \frac{2x^3+7x+5}{(x^2+x+2)(x^2+1)}$

I have tried dividing first but keep running into problem after problem, please help.

Answer 1

try writing your fraction as a sum of $\frac{A_0+A_1 x}{x^2+1}+\frac{B_0+B_1 x}{x^2+x+2}$ and then find $A_0,A_1$ and $B_0,B_1$ by equating $(A_0+A_1 x) (x^2+x+2)+(B_0+B_1 x)(x^2+1) = 2x^3+7x+5$

The final answer should be

$\frac{2x^3+7x+5}{(x^2+x+2)(x^2+1)}=\frac{5}{x^2+1}+\frac{2x-5}{2 + x + x^2}$

I hope that helps.

Answer 2

Step 1: Find $a, b, c, d$ such that: $\dfrac{2x^3 + 7x + 5}{(x^2 + x + 2)(x^2 + 1)} = \dfrac{ax + b}{x^2 + x + 2} + \dfrac{cx + d}{x^2 + 1}$

Step 2: Rewrite: $\dfrac{ax + b}{x^2 + x + 2} + \dfrac{cx+d}{x^2 + 1} = \dfrac{1}{2}\cdot \dfrac{a(2x+1)}{x^2 + x + 2} + \dfrac{1}{2}\cdot\dfrac{2b - a}{(x + \frac{1}{2})^2 + (\frac{\sqrt{7}}{2})^2} + \dfrac{c}{2}\cdot \dfrac{2x}{x^2 + 1} + d\cdot\dfrac{1}{x^2+1}$.

Step 3: Make the right substitution and complete answer.