# Thread: fraction decomposition of a rational function

1. ## fraction decomposition of a rational function

The form of the partial fraction decomposition of a rational function is given below.

A=
B=
C=

Now evaluate the indefinite integral.
i need a b and c and the integral

2. Originally Posted by dat1611
The form of the partial fraction decomposition of a rational function is given below.

A=
B=
C=

Now evaluate the indefinite integral.
i need a b and c and the integral
$\frac{{2{x^2} + 3x + 30}}
{{\left( {x + 4} \right)\left( {{x^2} + 9} \right)}} = \frac{A}
{{x + 4}} + \frac{{Bx + C}}
{{{x^2} + 9}} \Leftrightarrow$

$\Leftrightarrow 2{x^2} + 3x + 30 = \left( {{x^2} + 9} \right)A + \left( {x + 4} \right)\left( {Bx + C} \right) =$

$= A{x^2} + 9A + B{x^2} + Cx + 4Bx + 4C =$

$= \left( {A + B} \right){x^2} + \left( {4B + C} \right)x + 9A + 4C \Rightarrow$

$\Rightarrow \left\{ \begin{gathered}
A + B = 2, \hfill \\
4B + C = 3, \hfill \\
9A + 4C = 30; \hfill \\
\end{gathered} \right.$
$\Leftrightarrow \left\{ \begin{gathered}A = 2 - B, \hfill \\
C = 3 - 4B, \hfill \\
9\left( {2 - B} \right) + 4\left( {3 - 4B} \right) = 30; \hfill \\
\end{gathered} \right. \Leftrightarrow$

$\Leftrightarrow \left\{ \begin{gathered}A = 2 - B, \hfill \\C = 3 - 4B, \hfill \\30 - 25B = 30; \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}A = 2, \hfill \\B = 0, \hfill \\C = 3. \hfill \\ \end{gathered} \right.$

$\frac{{2{x^2} + 3x + 30}}
{{\left( {x + 4} \right)\left( {{x^2} + 9} \right)}} = \frac{2}
{{x + 4}} + \frac{3}
{{{x^2} + 9}}.$

3. Here is how I would do it...

1) Multiply both sides by $x+4$ and take the limit $\lim_{x\rightarrow{-4}}x+4$. This will automatically give you $A=2$

2) Multiply both sides by $x$ and take the limit as x goes to infinity. You will find $B=0$.

3) Simply, let $x=0$ and you will find $C=3$ right away.

So, $A=2, B=0,$ and $C=3$.

Thats it!

4. Sorry about the double post!

5. Originally Posted by dat1611

Now evaluate the indefinite integral.
i need a b and c and the integral
$\int {\frac{{2{x^2} + 3x + 30}}{{\left( {x + 4} \right)\left( {{x^2} + 9} \right)}}dx} = 2\int {\frac{{dx}}{{x + 4}}} + 3\int {\frac{{dx}}
{{{x^2} + 9}}} .$

$2\int {\frac{{dx}}{{x + 4}}} = 2\int {\frac{{d\left( {x + 4} \right)}}
{{x + 4}}} = 2\ln \left| {x + 4} \right| + C.$

$3\int {\frac{{dx}}{{{x^2} + 9}}} = \frac{1}{3}\int {\frac{{dx}}{{{{\left( {x/3} \right)}^2} + 1}}} = \left\{ \begin{gathered}\frac{x}{3} = u, \hfill \\\frac{1}{3}dx = du \hfill \\ \end{gathered} \right\} =$

$= \int {\frac{{dx}}{{{u^2} + 1}}} = \arctan u + C = \arctan \frac{x}{3} + C.$

$\boxed{\int {\frac{{2{x^2} + 3x + 30}}{{\left( {x + 4} \right)\left( {{x^2} + 9} \right)}}dx} = 2\ln \left| {x + 4} \right| + \arctan \frac{x}
{3} + C}$