Answer
$\dfrac{x}{(x+1)(x^2+9)}=-\dfrac{1}{10(x+1)}+\dfrac{x+9}{10(x^2+9)}$
Work Step by Step
We are given the fraction:
$\dfrac{x}{(x+1)(x^2+9)}$
As the denominator is already factored, we can write the partial fraction decomposition:
$\dfrac{x}{(x+1)(x^2+9)}=\dfrac{A}{x+1}+\dfrac{Bx+C}{x^2+9}$
Multiply all terms by the least common denominator $(x+1)(x^2+9)$:
$(x+1)(x^2+9)\cdot\dfrac{x}{(x+1)(x^2+9)}=(x+1)(x^2+9)\cdot\dfrac{A}{x+1}+(x+1)(x^2+9)\cdot\dfrac{Bx+C}{x^2+9}$
$x=A(x^2+9)+(Bx+C)(x+1)$
$x=Ax^2+9A+Bx^2+Bx+Cx+C$
$x=(A+B)x^2+(B+C)x+(9A+C)$
Identify the coefficients and build the system of equations:
$\begin{cases}
A+B=0\\
B+C=1\\
9A+C=0
\end{cases}$
Solve the system: multiply Equation 2 by -1 and add it to Equation 3:
$\begin{cases}
A+B=0\\
-B-C=-1\\
9A+C=0
\end{cases}$
$\begin{cases}
A+B=0\\
9A+C-B-C=0-1
\end{cases}$
$\begin{cases}
A+B=0\\
9A-B=-1
\end{cases}$
Add the two equations:
$A+B+9A-B=0-1$
$10A=-1$
$A=-\dfrac{1}{10}$
$A+B=0$
$-\dfrac{1}{10}+B=0$
$B=\dfrac{1}{10}$
$B+C=1$
$\dfrac{1}{10}+C=1$
$C=1-\dfrac{1}{10}$
$C=\dfrac{9}{10}$
The partial fraction decomposition is:
$\dfrac{x}{(x+1)(x^2+9)}=-\dfrac{\dfrac{1}{10}}{x+1}+\dfrac{\dfrac{1}{10}x+\dfrac{9}{10}}{x^2+9}$
$\dfrac{x}{(x+1)(x^2+9)}=-\dfrac{1}{10(x+1)}+\dfrac{x+9}{10(x^2+9)}$
