Partial Fractions

[goku900]

thanks

 

[Deleted member 4993]

I have the integral of (x^3)/(x^2+4x+3)dx from 0 to 1.

I did long division because the numerator degree was larger than the denominator and found that View attachment 2710

Now how do I break this up for my partial fraction decomposition?

Can you factorize the denominator?

 

[goku900]

yeah it factors to (x+1)(x+3) I know that. I'm just unsure which formula to use in my numerator

Does this look right

13x+12/(x+1)(x+3) = A/(x+1) +B/(x+3)

13x+12= A(x+3) + B(x+1)

After solving for B and A (let x=-1 to solve for A, let x=-3 to solve for B)

I get A = -11/2
I get B = 37/2

From there it becomes

x^3/(x+1)(x+3) = x-4 +(11/2)/(x+1) + (37/2)/(x+3)

From here, Do I evaluate this as three integrals and add them ?

 

[Deleted member 4993]

yeah it factors to (x+1)(x+3) I know that. I'm just unsure which formula to use in my numerator

Does this look right

13x+12/(x+1)(x+3) = A/(x+1) +B/(x+3)

13x+12= A(x+3) + B(x+1)

After solving for B and A (let x=-1 to solve for A, let x=-3 to solve for B)

I get A = -11/2
I get B = 37/2

From there it becomes

x^3/(x+1)(x+3) = x-4 +(11/2)/(x+1) + (37/2)/(x+3)

From here, Do I evaluate this as four, namely, [x, -4, (11/2)/(x+1) & (37/2)/(x+3)] integrals and add them ? .................Yes ............ Why do you have doubt/s about that?

.

 

[goku900]

I need to evaluate them now from lower bound 0 to upper bound 1. My calculator says the answer is .037 but when I do it out I'm getting like -3 something

 

[Deleted member 4993]

I need to evaluate them now from lower bound 0 to upper bound 1. My calculator says the answer is .037 but when I do it out I'm getting like -3 something

Correct answer is ~ 0.037

 

[DrPhil]

yeah it factors to (x+1)(x+3) I know that. I'm just unsure which formula to use in my numerator

Does this look right

13x+12/(x+1)(x+3) = A/(x+1) +B/(x+3)

13x+12= A(x+3) + B(x+1)

After solving for B and A (let x=-1 to solve for A, let x=-3 to solve for B)

I get A = -11/2 ?
I get B = 37/2

From there it becomes

x^3/(x+1)(x+3) = x-4 +(11/2)/(x+1) + (37/2)/(x+3)

From here, Do I evaluate this as three integrals and add them ?

"let x=-1 to solve for A" --> -13 + 12 = A(-1 + 3) --> A = -1/2
"let x=-3 to solve for B" --> -39 + 12 = B(-3 + 1) --> B = 27/2

x^3/(x+1)(x+3) = x - 4 - (1/2)/(x+1) + (27/2)/(x+3)

Yes - evaluate the four integrals and add them. Recalculate the integrals with the correct A and B, and you should get the correct answer of 0.037 - or at least I did!

 

[lookagain]

Does this look right

13x+12/[(x+1)(x+3)] . . . . You need grouping symbols around the denominator.


= A/(x+1) +B/(x+3)

13x+12= A(x+3) + B(x+1)

After solving for B and A (let x=-1 to solve for A, let x=-3 to solve for B)

I get A = -11/2
I get B = 37/2

From there it becomes

x^3/[(x+1)(x+3)] = . . . . Here also.


x-4 +(11/2)/(x+1) + (37/2)/(x+3)

From here, Do I evaluate this as three integrals and add them ?

- - - - - - - - - - - - - - - -

x^3/[(x+1)(x+3)]= And they are needed here, also.


x - 4 - (1/2)/(x+1) + (27/2)/(x+3)