Question

integral of (3x+1) / ((x^3)(3x+2)(x-1)) dx

Answer 1

\[\int\limits \frac{ 3x+1 }{ x^3 (3x+2)(x-1)}\]

Answer 2

I think you use partial fractions here so.
\[3x+1=A(x^5)(3x+2)(x-1)+B(x^4)(3x+2)(x-1)\]
\[+C(x^3)(3x+2)(x-1)+D(x^6)(x-1)+E(x^6)(3x+2)\]

Answer 3

using x=1 I was able to get E=4/5
what other numbers for x should I use, did I type this correctly?

Answer 4

\[\frac{ 3x+1 }{ x^3(3x+2)(x-1) }=\frac{ A }{ x }+\frac{ B }{ x^{2} }+\frac{ C }{ x^{3} }+\frac{ D }{ 3x+2 }+\frac{ E }{ x-1 }\]

Answer 5

\[3x+1 = Ax^2(3x+2)(x-1) + Bx(3x+2)(x-1) + C(3x+2)(x-1) + Dx^3(x-1) + Ex^3(3x+2)\]

Answer 6

for A wouldn't you have to use the x^3 though?

Answer 7

oh, nevermind. I understand XD

Answer 8

The cut off part:

\[Dx^3(x-1) + Ex^3(3x+2)\]

No, you do not. You have 5 linear factors on bottom, x*x*x*(3x+2)(x-1). For each of those linear factors, you only need a constant in the each of their fractions for the partial fraction decomp.

Answer 9

at least that explains the weird answers part XD

Answer 10

Yeah, you would only need an x^3 on top if you had a factor that was some sort of quartic factor like x^4 + 3x^2 + 1 or something silly like that, lol.

Answer 11

what numbers for x would you suggest

Answer 12

0 gave me C=1/2

Answer 13

what does 1 give then?

Answer 14

E=4/5

Answer 15

and then -2/3 for x

Answer 16

And C should be -1/2, not just 1/2 :)

Answer 17

D=-81/20

Answer 18

Sould be -81/40

Answer 19

haha I meant to type that XD

Answer 20

So what would we use for A&B since we used up our roots

Answer 21

We can pick anything we want. Something that will make calculations as easy as possible. In the end, we will need to do a system of equations for A and B. Let's try -1 first and see what we get left with.

Answer 22

25/8=A+B.... That doesn't look right...

Answer 23

It still might be. We need a second equation with A and B so we can do a system of equations. So we can pick another x value, like x = 2, and see what we get. Then take those 2 equations to solve for A and B. If the algebra goes wrong and we have issues wih that, there's another way to solve for A and B without picking x-values.

Answer 24

how would we do that?

Answer 25

I just need to solve A and B then I know how to do the other stuff