# Thread: Partial fractions, indefinte integral

1. ## Partial fractions, indefinte integral

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

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

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

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

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

4. Hello, matt3D!

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

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

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 . . .

5. 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. ^^'

6. 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.

7. 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?