Evaluate int (3x+3)/(x^3-1) dx

[confused_07]

Evaluate int (3x+3)/(x^3-1) dx

I believe I am supposed to use the "integration by Partial fractions" Methos. In that case, I get the denominator to break down to:

(x-1)(x^2+x+1)

So, my equation should be:

[(3x+3)/(x^3-1)] = {(A/x) + [(B/(x-1)] + [(Cx+D)/(x^2+x+1)]}

Multiply both sides by (x^3-1) yields:

(3x+3) = A(x^2-1) + B(x^2+x+1) + (Cx+D)(x-1)
(3x+3) = (Ax^2 - A) + (Bx^2+Bx+B) + (Cx^2-C+Dx-D)

Gather like coefficients:

(3x+3) = [(A+B+C)x^2] + [(B+D-C)x] + (B - D - A)

From here things don't come out right. Did I factor correctly? If so, did I do the rest of it right?

Thanks for taking the time to read all that....

 

[soroban]

Hello, confused_07!

Your set-up is wrong . . .

Integrate: \(\displaystyle \L\L\int\frac{3x\,+\,3}{x^3\,-\,1}\,dx\)

I believe I am supposed to use the "integration by Partial fractions" Method . . . Yes!

In that case, I get the denominator to break down to: \(\displaystyle \x\,-\,1)(x^2\,+\,x\,+\,1)\;\) . . . Right!

So, my equation should be: \(\displaystyle \L\:\frac{3x\,+\,3}{x^3\,-\,1}\;=\;\frac{A}{x}\,+\,\frac{B}{x\,-\,1}\,+\,\frac{Cx\,+\,D}{x^2\,+\,x\,+\,1}\;\) . . . no
. . . . . . . . . . . . . . . . . . . . . . . . - . . . . . ?

It should be: \(\displaystyle \L\:\frac{3x\,+\,3}{x^3\,-\,1}\;=\;\frac{A}{x\,-\,1}\,+\,\frac{Bx}{x^2\,+\,x\,+\,1}\,+\,\frac{C}{x^2\,+\,x\,+\,1}\)

Then: \(\displaystyle \:3x\,+\,3\;=\;(A\,+\,B)x^2\,+\,(A\,-\,B\,\,+\,C)x\,+\,(A\,-\,C)\)

Got it?

 

[confused_07]

I kind of see what you did, but my text book always pulls a "A/x" out of the equations. Why did you use the (x^2+x+1) twice?

 

[soroban]

Hello, confused_07!

I kind of see what you did
. Upi "kind of see?" . . . What rules were you taught?

but my text book always pulls a "A/x" out of the equations.
Always? . . . That can't be right!

Why did you use the (x^2+x+1) twice?

I hoped that it was obvious . . .

Most people were taught to write: . . . \(\displaystyle \L\frac{A}{x}\;\;+\;\;\frac{Bx\,+\,C}{x^2\,+\,x\,+\,1}\)

I make two fractions immediately: \(\displaystyle \L\:\frac{A}{x}\,+\,\overbrace{\frac{Bx}{x^2\,+\,x\,+\,1}\,+\,\frac{C}{x^2\,+\,x\,+\,1}}\)

Why? . It saves a lot of extra algebra later.

 

[confused_07]

Thanks for the answer, I understand it now. I say always as every example it gave had an "A/x" pulled out. Unfortunately, I am taking a self-taught course, so the only way I am taught is how I read the text. Thanks again!