Wondering how these can be equal:

[Vay]

-(23/27) * integral of (x+6)/( (x^2)+3x+9 ) dx = -(23/27) * integral of (x+(3/2))/( (x^2)+3x+9 ) dx - (23/27) * integral of (9/2)/((x+(3/2))^2) + (27/4)). I don't know how this is done, primarily, everything else is the same, except for the numerator, where the left hand side is (x+6), which is equal to the right hand side of (x+(3/2)-(9/2)). How is this possible?

Edited September 17, 2012 by Vay

[Orion1]

Your equation as stated is not an identity, it is a false statement.

 

Perhaps someone else here can verify my claim.

Edited September 17, 2012 by Orion1

[imatfaal]

[math] -\frac{23}{27} \int{\frac{x+6}{x^2+3x+9}} dx = -\frac{23}{27} \int{\frac{x+3/2}{x^2+3x+9}} dx - \frac{23}{27} \int{\frac{9/2}{(x+3/2)^2+ \frac{27}{4}}}dx [/math]

 

Vay

is that what you want to show? Your last section has 5 opening brackets and 6 closing!

 

Lose the funky -23/27 and tidy

 

[math] \int{\frac{x+6}{x^2+3x+9}} dx = \int{\frac{x+3/2}{x^2+3x+9}} dx - \int{\frac{9/2}{x^2+3x+9}}dx [/math]

 

and work out the basis integrals to check that the addition of numerators is acceptable. you might be pleasantly surprised. I have never really liked integration and thus I am at a loss as to the constants! You have three constants with indefinite integration and I see no real reason why they should cancel - perhaps they do, perhaps they don't but you would need to find a more knowledgable poster than me to explain other than the obvious! BTW in adding up the numerators you have got confused with the minus signs you should have -(x+6) and -(x+3/2)-(9/2) which do work out the same

[Vay]

I have the pdf of the solution where I got it from (webassign calculus class if you know what that is). This was the solution for the topic I am learning on integration by partial fractions etc. The equation of integrals is near the middle as you can see.

1eac7657014f7464cd701d8473796424.pdf

Edited September 17, 2012 by Vay

[Orion1]

Problem:

Integrate:

[math]\int \frac{1}{x^3 - 27} dx[/math]

 

The first step involves factoring and splitting the denominator with some theorem:

[math]\frac{1}{x^3 - 27} = \frac{1}{(x - 3)(x^2 + 3x + 9)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 3x + 9}[/math]

What is the name of this theorem?

 

The next step involves normalizing the theorem:

[math]1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)[/math]

How was the coefficients of B and C derived and solved?

 

The solution only states that:

[math]A = \frac{1}{27}[/math]

 

[math]0 = \frac{1}{27} + B[/math]

 

[math]1 = \frac{1}{3} + 3C[/math]

 

And:

[math]B = -\frac{1}{27}[/math] and [math]C = -\frac{2}{9}[/math]

 

I can see how the [math]A[/math] coefficient was derived from the x-axis zero intercept, however, it is not entirely clear to me how these other coefficients were derived and solved?

 

The next step after solving the coefficients listed in that equation solution involves multiplying completely through by a constant and a numerical integration of coefficients via substitution:

[math]\int \frac{23}{x^3 - 27} dx = 23 \int \frac{\frac{1}{27}}{x - 3} dx + 23 \int \frac{-\frac{1}{27}x - \frac{2}{9}}{x^2 + 3x + 9} dx[/math]

 

However there is a missing first principle as to why this particular constant '23' was used without prior knowledge what the solution is. This equation solution has been reverse engineered from a known solution.

The equation also requires the memorization of these identities:

[math]\int \frac{1}{x \pm b} dx = \ln (x \pm b) + C[/math]

 

[math]\int \frac{x + c_1}{x^2 + b_2 x + c_2} dx = \frac{1}{2} \log \left(x^2 + b_2 x + c_2 \right) - \frac{\left(b_2 - 2 c_1 \right) \tan ^{-1} \left(\frac{2 x + b_2}{\sqrt{4 c_2 - b_2^2}} \right)}{\sqrt{4 c_2 - b_2^2}} + C[/math]

 

The solution is requiring to factor and split an indefinite integral, however there is a missing first principle as to which direction to factor into what identity.

 

I fail to see how this equation can be effectively solved without a comprehensive identities table hard copy and/or math software.

Reference

Derivation of equation and solution

Edited September 19, 2012 by Orion1

[Orion1]

Solving the coefficients:

Given:

[math]1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)[/math]

 

Set x = 3 to zero out the (Bx + C) term and numerically integrate via substitution:

[math]1 = A(3^2 + 3(3) + 9) + (B(3) + C)(3 - 3) = A(27)[/math]

[math]1 = A(27)[/math]

 

Solve for A:

[math]\boxed{A = \frac{1}{27}}[/math]

 

Set x = 0 to zero out the Bx term and numerically integrate via substitution:

[math]1 = \frac{(0^2 + 3(0) + 9)}{27} + (B(0) + C)(0 - 3) = \frac{9}{27} - 3C[/math]

[math]1 = \frac{9}{27} - 3C[/math]

[math]-\frac{27}{27} + \frac{9}{27} = -\frac{18}{27} = -\frac{2}{3} = 3C[/math]

 

Solve for C:

[math]\boxed{C = -\frac{2}{9}}[/math]

 

Set x = 1 as arbitrary and numerically integrate via substitution:

[math]1 = \frac{(1^2 + 3(1) + 9)}{27} + \left(B(1) - \frac{2}{9} \right)(1 - 3) = \frac{13}{27} - 2B + \frac{3}{3} \times \frac{4}{9} = \frac{13}{27} - 2B + \frac{12}{27} = \frac{25}{27} - 2B[/math]

[math]1 = \frac{25}{27} - 2B[/math]

[math]-\frac{27}{27} + \frac{25}{27} = -\frac{2}{27} = 2B[/math]

 

Solve for B:

[math]\boxed{B = -\frac{1}{27}}[/math]