242t.8.5.9 expand the quotient by partial fractions

[karush]

$\tiny{242t.8.5.9}$
$\textsf{expand the quotient by}$ $\textbf{ partial fractions}$
\begin{align*}\displaystyle
y&=\int\frac{dx}{9-25x^2} &\tiny{(1)}\\
\end{align*}
$\textit{expand and multiply every term by $(3+5x)(3-5x)$}$
\begin{align*}\displaystyle
\frac{1}{9-25x^2}&=\frac{A}{(3-5x)}-\frac{B}{(3+5x)}\\
1&=A(3+5x)-B(3-5x)
\end{align*}
$\textit{ok my question here is.. if you use values to get A0 or B0 you have to use fractions}\\$
$\textit{ is there alt method or just go with $x=\pm \frac{3}{5}$}\\ \\$
$\textit{book answer}$
\begin{align*}\displaystyle
y_{pf}&=\frac{1}{30}(\ln\left|5x+3 \right|-\ln|5x-3|)+C
\end{align*}

 

[topsquark]

$\tiny{242t.8.5.9}$
$\textsf{expand the quotient by}$ $\textbf{ partial fractions}$
\begin{align*}\displaystyle
y&=\int\frac{dx}{9-25x^2} &\tiny{(1)}\\
\end{align*}
$\textit{expand and multiply every term by $(3+5x)(3-5x)$}$
\begin{align*}\displaystyle
\frac{1}{9-25x^2}&=\frac{A}{(3-5x)}-\frac{B}{(3+5x)}\\
1&=A(3+5x)-B(3-5x)
\end{align*}
$\textit{ok my question here is.. if you use values to get A0 or B0 you have to use fractions}\\$
$\textit{ is there alt method or just go with $x=\pm \frac{3}{5}$}\\ \\$
$\textit{book answer}$
\begin{align*}\displaystyle
y_{pf}&=\frac{1}{30}(\ln\left|5x+3 \right|-\ln|5x-3|)+C
\end{align*}

You aren't setting x to be anything. You have two equations:
1 = 3A - 3B
0 = 5Ax + 5Bx
which has to be satisfied by all x. So the second equation becomes
0 = 5A + 5B

So the second equation gives B = -A, so putting this into the first equation:
1 = 3A - 3(-A) = 6A

or A = 1/6 and B = -1/6.

Your integral then becomes
\(\displaystyle \int \frac{dx}{9 - 25x^2} = \int \left ( \frac{1/6}{3 - 5x} + \frac{-1/6}{3 + 5x} \right ) ~dx\)

and now you can integrate.

-Dan

 

[karush]

here is what I did but your method better...

expand and multiply every term by $(3+5x)(3-5x)$}
\begin{align*}\displaystyle
\frac{1}{9-25x^2}&=\frac{A}{(3-5x)}-\frac{B}{(3+5x)}\\
1&=A(3+5x)-B(3-5x)
\end{align*}
$\textit{if $\displaystyle x=\frac{3}{5}$}$
\begin{align*}\displaystyle
1&=6A \therefore A=\frac{1}{6}
\end{align*}
$\textit{if $\displaystyle x=-\frac{3}{5}$}$
\begin{align*}\displaystyle
1&=-6B \therefore B=-\frac{1}{6}
\end{align*}
Integrate and simplify
\begin{align*}\displaystyle
I_{9}&=\frac{1}{6}\int\frac{1}{(3+5x)}
-\frac{1}{6}\int\frac{1}{(3+5x)} \\
&=\color{red}{
\frac{1}{30}(\ln\left|5x+3 \right|-\ln|5x-3|)+C}
\end{align*}

 

[topsquark]

...if x=3/5...

Okay, I see what you are doing now. Yes, you can do it but I find it to be a bit confusing since you are (temporarily) giving x a value. Since you can use any value for x you can get away with it. But it may wind up causing some confusion.

I don't think I've seen this way of doing it.

-Dan