# Find the partial fractions decomposition and an antiderivative. \frac{3x + 8}{x^3 + 5x^2 + 6x}

## Question:

Find the partial fractions decomposition and an antiderivative.

{eq}\frac{3x + 8}{x^3 + 5x^2 + 6x} {/eq}

## Partial Fractions Decomposition:

We can write {eq}\frac{3x + 8}{x^3 + 5x^2 + 6x}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x+3} {/eq}

We know that {eq}\int \frac{1}{x}\,dx=\log x {/eq}

Partial Fraction Decomposition:

{eq}x^3 + 5x^2 + 6x=x\left ( x^2 + 5x + 6 \right )=x\left ( x^2 + 3x+2x + 6 \right )=x\left [ x\left ( x+3 \right ) +2(x+3)\right ]=x(x+2)(x+3) {/eq}

Therefore,

{eq}\frac{3x + 8}{x^3 + 5x^2 + 6x}=\frac{3x + 8}{x(x+2)(x+3)}=\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x+3}\\ \Rightarrow 3x+8=A(x+2)(x+3)+Bx(x+3)+Cx(x+2)\\ x=0\,;8=6A\Rightarrow A=\frac{4}{3}\\ x=-2\,;2=-2B\Rightarrow B=-1\\ x=-3\,;-1=3C\Rightarrow C=\frac{-1}{3}\\ \therefore \frac{3x + 8}{x^3 + 5x^2 + 6x}=\frac{\frac{4}{3}}{x}-\frac{1}{x+2}+\frac{\frac{-1}{3}}{x+3} {/eq}

Anitiderivative:

{eq}\int \frac{3x + 8}{x^3 + 5x^2 + 6x}\,dx=\int \frac{\frac{4}{3}}{x}-\frac{1}{x+2}+\frac{\frac{-1}{3}}{x+3}\,dx\\ =\frac{4}{3}\log x-\log (x+2)-\frac{1}{3}\log (x+3)+C {/eq}