Partial fractions, indefinte integral

**matt3D** · September 16th 2008, 09:53 AM

Hi, I have the problem:

and it states that the integrand has partial fractions decomposition:

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

**Moo** · September 16th 2008, 10:11 AM

Hello,

Originally Posted by matt3D

Hi, I have the problem:

and it states that the integrand has partial fractions decomposition:

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 $x^2(x^2+25)$

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

Please spot out your mistakes...

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

**11rdc11** · September 16th 2008, 11:06 AM

A=3
B=5
C=2
D=2

**Soroban** · September 16th 2008, 11:32 AM

Hello, matt3D!

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

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

Note that I immediately made two fractions for the quadratic.
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]: . $2D \:=\:4 \quad\Rightarrow\quad\boxed{ D \:=\:2}$

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

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 . . .

**Moo** · September 16th 2008, 11:37 AM

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

And this teacher went further than this by taking the derivative when we had something like x and x² in the partial fractions decomposition. ^^'

**matt3D** · September 16th 2008, 04:12 PM

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 $x^2(x^2+25)$

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

Please spot out your mistakes...

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

Yea, I was super tired when I was doing this.

**matt3D** · September 16th 2008, 04:14 PM

Originally Posted by Soroban

Hello, matt3D!

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

Note that I immediately made two fractions for the quadratic.
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]: . $2D \:=\:4 \quad\Rightarrow\quad\boxed{ D \:=\:2}$

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

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?

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