i know this really isnt calculus. but we are doing intergration and are running into problems like this...

Using partial fractions can someone show me how you would solve this please?

And also when dividing functions i really forgot how to do it.. can sum1 show me the work/how they would divide this please?

2. Hello,

For the 1st one, I don't see where partial fractions are...

For the second one, is it $\frac{x^2}{x^3+4}$ ? What do you want exactly?

3. Originally Posted by Moo
Hello,

For the 1st one, I don't see where partial fractions are...

For the second one, is it $\frac{x^2}{x^3+4}$ ? What do you want exactly?
the first one is partial fractions this is as far as i can get i really need help

the second one i want to divide the fraction.. u know kinda like synthetic division... but iforgot how to do it...

4. Ok for the first one.
I don't understand a thing, is it Bx or what ? Because you replaced it by Bx²+Bx...

For the second one, you can't go further since the degree of the upper polynomial is inferior to the degree of the lower polynomial..

5. Originally Posted by Moo
Ok for the first one.
I don't understand a thing, is it Bx or what ? Because you replaced it by Bx²+Bx...

For the second one, you can't go further since the degree of the upper polynomial is inferior to the degree of the lower polynomial..

this is the un multiplied one (forgot a Parentheses around the bx+c)

then here i multiplied it out (forgot to add a "A" when multiplying the 9)

6. 1. What was the original fraction that you were trying to integrate?

2. So I'm guessing your original fraction was this then: $\frac{x^{3} + 4}{x^{2}}$ seeing how the $x^{2}$ is the divisor in your image. Personally, I'd just split the fraction up: $\frac{x^{3} + 4}{x^{2}} = \frac{x^{3}}{x^{2}} + \frac{4}{x^{2}} = x + \frac{4}{x^{2}}$

7. Ok, so you've got actually 3 equations

As $10=0x^2+0x+10$, you should have :

$A+B=0$
$-B+C=0$
$9-C=10$

8. Originally Posted by Moo
Ok, so you've got actually 3 equations

As $10=0x^2+0x+10$, you should have :

$A+B=0$
$-B+C=0$
$9-C=10$

$A+B=10$
$-B+C=0$
$9A-C=0$

now i have to find the values for a,b and c.. and this is where i am having difficulties..

9. Originally Posted by o_O
1. What was the original fraction that you were trying to integrate?

2. So I'm guessing your original fraction was this then: $\frac{x^{3} + 4}{x^{2}}$ seeing how the $x^{2}$ is the divisor in your image. Personally, I'd just split the fraction up: $\frac{x^{3} + 4}{x^{2}} = \frac{x^{3}}{x^{2}} + \frac{4}{x^{2}} = x + \frac{4}{x^{2}}$

oh ok nice method that way.. but i think i really need to know how to do it without spliting it up.. can u explain to me that method

10. Originally Posted by Legendsn3verdie
$A+B=10$
$-B+C=0$
$9A-C=0$

now i have to find the values for a,b and c.. and this is where i am having difficulties..
Is it 10 or 10x² in the LHS ?

So let's suppose that your equations are correct...
To find the values for A, B and C, use substitutions !

From the second equation, you have B=C.

From the first equation : A=10-B.

Replacing in the third equation : 9A-C=9(10-B)-B=90-10B=0

Hence, B=...

11. Originally Posted by o_O
1. What was the original fraction that you were trying to integrate?

2. So I'm guessing your original fraction was this then: $\frac{x^{3} + 4}{x^{2}}$ seeing how the $x^{2}$ is the divisor in your image. Personally, I'd just split the fraction up: $\frac{x^{3} + 4}{x^{2}} = \frac{x^{3}}{x^{2}} + \frac{4}{x^{2}} = x + \frac{4}{x^{2}}$

the original problem i was trying to intergrate for the first this was...

$\int \frac{10}{(x-1)(x^{2}+9)}$

12. Originally Posted by Legendsn3verdie
the original problem i was trying to intergrate for the first this was...

$\int \frac{10}{(x-1)(x^{2}+9)}$
So it's

$\frac{A}{x-1}+\frac{Bx+C}{x^2+9}$

A(x²+9)+(Bx+C)(x-1)=10=10+0x+0x²

$Ax^2+9A+Bx^2+Cx-Bx-C=0x^2+0x+10$

$x^2(A+B)+(-B+C)x+(9A-C)=0x^2+0x+10$

Thus :
A+B=0
-B+C=0
9A-C=10

Not like you said... ^^

13. From the second equation, you get B=C.

Replacing in the first equation, A+B=0 <=> A=-C

Replacing in the third equation, 9A-C=10 <=> -9C-C=10 <=> C=...

14. Originally Posted by Moo
So it's

$\frac{A}{x-1}+\frac{Bx+C}{x^2+9}$

A(x²+9)+(Bx+C)(x-1)=10=10+0x+0x²

$Ax^2+9A+Bx^2+Cx-Bx-C=0x^2+0x+10$

$x^2(A+B)+(-B+C)x+(9A-C)=0x^2+0x+10$

Thus :
A+B=0
-B+C=0
9A-C=10

Not like you said... ^^

hm ahh i see what u did there.. deffintly helped me.. can u show me how u'd solve for abc now.

15. Here's a link on how to do long division: Polynomial Long Division

As for your example, try to see if you can link this to what they have:

$\begin{array}{lllll} \quad \:\: \: x \: \: {\color{red} + \: \: \: \: 0} \: \: \: {\color{blue} + \: \: 0 \:} {\color{magenta} \: + \: 0} \\ x^{2} \overline{| \:x^{3} + 0x^{2} + 0x + 4} \\ \: \: - \underline{\: x^{3} } \\ \quad \: \: \: 0 \: + 0x^{2} \\ \: \: \quad - \underline{ \quad \: \: {\color{red}0x^{2}}} \end{array}$
$\begin{array}{lllll} {\color{white} .} \qquad \qquad 0 \: + 0x \\ \qquad \quad \: \: - \underline{ \qquad \: {\color{blue} 0} \:} \\ \qquad \qquad \qquad \: \: 0 \: + \: 4 \\ \qquad \qquad \quad \: \: \: - \underline{ \qquad \: {\color{magenta} 0} \:} \\ \qquad \qquad \qquad \qquad \: \: \: {\color [cmyk] {0.75,0,0.75,0.45}4} \end{array}$

Put the remainder 4 on top of the divisor and we have: $\frac{x^{3} + 4}{x^{2}} = x + {\color [cmyk] {0.75,0,0.75,0.45}\frac{4}{x^{2}}}$