# Partial fractions, indefinte integral

• Sep 16th 2008, 10:53 AM
matt3D
Partial fractions, indefinte integral
Hi, I have the problem:
http://webwork.csufresno.edu/webwork...54eb9ae4c1.png
and it states that the integrand has partial fractions decomposition:
http://webwork.csufresno.edu/webwork...836e2f37a1.png
So then I need to make $7x^3+5x^2+125x+75=a(x)(x^2+25)+b(x^2)(cx+d)+(cx+d) (x^3)$
Now, would ax^3=7? I know what to do with simple A and B partial fractions but this is looking difficult especially when there is multiple variables multiplied by each other. How can I solve this?
Thanks,
Matt
• Sep 16th 2008, 11:11 AM
Moo
Hello,
Quote:

Originally Posted by matt3D
Hi, I have the problem:
http://webwork.csufresno.edu/webwork...54eb9ae4c1.png
and it states that the integrand has partial fractions decomposition:
http://webwork.csufresno.edu/webwork...836e2f37a1.png
So then I need to make $7x^3+5x^2+125x+75=a(x)(x^2+25)+b(x^2)(cx+d)+(cx+d) (x^3)$
Now, would ax^3=7? I know what to do with simple A and B partial fractions but this is looking difficult especially when there is multiple variables multiplied by each other. How can I solve this?
Thanks,
Matt

Nope, you have to develop completely (and correctly lol)

$\frac{a}{x^2}+\frac bx+\frac{cx+d}{x^2+25}$
The common denominator is $x^2(x^2+25)$

$\implies =\frac{a(x^2+25)+b(x)(x^2+25)+(cx+d)(x^2)}{x^2(x^2 +25)}$

$a(x^2+25)+b(x)(x^2+25)+(cx+d)(x^2)=\dots=x^3(b+c)+ x^2(a+d)+x(25b)+25(a)$
• Sep 16th 2008, 12:06 PM
11rdc11
A=3
B=5
C=2
D=2
• Sep 16th 2008, 12:32 PM
Soroban
Hello, matt3D!

Quote:

$\int\frac{7x^3 + 5x^2 + 125x + 75}{x^4+25x^2}\,dx$

We have: . $\frac{7x^3 + 5x^2 + 125x + 75}{x^2(x^2+25)} \;=\;\frac{A}{x} + \frac{B}{x^2} + \frac{Cx}{x^2+25} + \frac{D}{x^2+25}$

I discovered this trick a half-century ago and have taught it that way ever since.
It seems that no teacher has discovered this yet!

Multiply through by the LCD:

. . $7x^3 + 5x^2 + 125x + 75 \;=\;x(x^2+16)\!\cdot\!A + (x^2+25)\!\cdot\!B + x^3\!\cdot\!C + x^2\!\cdot\!D$

Let $x = 0\!:\quad 75 \;=\;25B \quad\Rightarrow\quad\boxed{B \:=\:3}$

Let $x = 1\!:\quad 212 \:=\:26A + 26B + C + D \quad\Rightarrow\quad 26A + C + D \:=\:134\;\;{\color{blue}[1]}$

Let $x = \text{-}1\!:\quad \text{-}52 \:=\:-26A + 26B - C + D \quad\Rightarrow\quad \text{-}26A - C + D \:=\:\text{-}130\;\;{\color{blue}[2]}$

Let $x = 2\!:\quad 401 \:=\:58A + 29B + 8C + 4D \quad\Rightarrow\quad29A + 4C + 2D \:=\:157\;\;{\color{blue}[3]}$

Add [1] and [2]: . $2D \:=\:4 \quad\Rightarrow\quad\boxed{ D \:=\:2}$

$\begin{array}{cccc}\text{Then {\color{blue}[1]} becomes:} & 26A + C &=& 132 \\
\text{and {\color{blue}[3]} becomes:} & 29A + 4C &=& 153 \end{array}$

Solve this system: . $\boxed{A \:=\:5}\quad\boxed{ C \:=\:2}$

Therefore: . $\frac{7x^3 + 5x^2 + 125x + 75}{x^2(x^2+25)} \;=\;\frac{5}{x} + \frac{3}{x^2} + \frac{2x}{x^2+25} + \frac{2}{x^2+25}$

. . which can be readily integrated . . .

• Sep 16th 2008, 12:37 PM
Moo
Quote:

It seems that no teacher has discovered this yet!
Well, my teacher taught it to us... and I'm sure he's not the only one. Some people used it on MHF too (Worried)

And this teacher went further than this by taking the derivative when we had something like x and x² in the partial fractions decomposition. ^^'
• Sep 16th 2008, 05:12 PM
matt3D
Quote:

Originally Posted by Moo
Hello,

Nope, you have to develop completely (and correctly lol)

$\frac{a}{x^2}+\frac bx+\frac{cx+d}{x^2+25}$
The common denominator is $x^2(x^2+25)$

$\implies =\frac{a(x^2+25)+b(x)(x^2+25)+(cx+d)(x^2)}{x^2(x^2 +25)}$

$a(x^2+25)+b(x)(x^2+25)+(cx+d)(x^2)=\dots=x^3(b+c)+ x^2(a+d)+x(25b)+25(a)$

Yea, I was super tired when I was doing this. (Rofl)
• Sep 16th 2008, 05:14 PM
matt3D
Quote:

Originally Posted by Soroban
Hello, matt3D!

We have: . $\frac{7x^3 + 5x^2 + 125x + 75}{x^2(x^2+25)} \;=\;\frac{A}{x} + \frac{B}{x^2} + \frac{Cx}{x^2+25} + \frac{D}{x^2+25}$

I discovered this trick a half-century ago and have taught it that way ever since.
It seems that no teacher has discovered this yet!

Multiply through by the LCD:

. . $7x^3 + 5x^2 + 125x + 75 \;=\;x(x^2+16)\!\cdot\!A + (x^2+25)\!\cdot\!B + x^3\!\cdot\!C + x^2\!\cdot\!D$

Let $x = 0\!:\quad 75 \;=\;25B \quad\Rightarrow\quad\boxed{B \:=\:3}$

Let $x = 1\!:\quad 212 \:=\:26A + 26B + C + D \quad\Rightarrow\quad 26A + C + D \:=\:134\;\;{\color{blue}[1]}$

Let $x = \text{-}1\!:\quad \text{-}52 \:=\:-26A + 26B - C + D \quad\Rightarrow\quad \text{-}26A - C + D \:=\:\text{-}130\;\;{\color{blue}[2]}$

Let $x = 2\!:\quad 401 \:=\:58A + 29B + 8C + 4D \quad\Rightarrow\quad29A + 4C + 2D \:=\:157\;\;{\color{blue}[3]}$

Add [1] and [2]: . $2D \:=\:4 \quad\Rightarrow\quad\boxed{ D \:=\:2}$

$\begin{array}{cccc}\text{Then {\color{blue}[1]} becomes:} & 26A + C &=& 132 \\
\text{and {\color{blue}[3]} becomes:} & 29A + 4C &=& 153 \end{array}$

Solve this system: . $\boxed{A \:=\:5}\quad\boxed{ C \:=\:2}$

Therefore: . $\frac{7x^3 + 5x^2 + 125x + 75}{x^2(x^2+25)} \;=\;\frac{5}{x} + \frac{3}{x^2} + \frac{2x}{x^2+25} + \frac{2}{x^2+25}$

. . which can be readily integrated . . .

Hi Soroban, thanks for the explanation. But why do you multiply A by x(x^2+16) I don't see that in the fractions, and D by x^2 and C by x^3?