1. ## Partial fraction

There is a problem in the partial problem exercise of my math book that is giving me some trouble working out. Here is the problem: $\frac {x^2+x+1}{(x^2-1)(x^2+1)}$ so I started break it up into smaller fraction $\frac {x^2+x+1}{(x^2-1)(x^2+1)}= \frac{Ax+C}{x^2-1}+ \frac {Bx+D}{x^2+1}$
$x^2+x+1 = (Ax+C)(x^2+1)+(Bx^2+D)(x^2-1)$ and here is where I got stuck. Can someone Please help me with this?

2. You have a mistake in your last line.
It should be
$(Ax+C)(x^2+1)+(Bx+D)(x^2-1)=x^2+x+1$

From here, expand your brackets, and then compare coefficients for each x term, ie $x^3, x^2, x$ and $x^0$

You should get
$A=0.5, B=-0.5, C=1, D=0$

3. Originally Posted by scrible
There is a problem in the partial problem exercise of my math book that is giving me some trouble working out. Here is the problem: $\frac {x^2+x+1}{(x^2-1)(x^2+1)}$

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

$\frac {x^2+x+1}{(x-1)(x+1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}$

$x^2+x+1 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx+D)(x^2-1)$

start your solution for A, B, C, and D

4. Would it be helpful to reduce the denominator further?

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

5. Originally Posted by I-Think
You have a mistake in your last line.
It should be
$(Ax+C)(x^2+1)+(Bx+D)(x^2-1)=x^2+x+1$

From here, expand your brackets, and then compare coefficients for each x term, ie $x^3, x^2, x$ and $x^0$

You should get
$A=0.5, B=-0.5, C=1, D=0$

thank you very much for helping me. I worked out the problem by expanding it and comparing the coefficient and thus far I got A+B=0, C+D=1, A-B=1,C-D=1 could you show me how do you come up with A=0.5, B=-0.5,C=1 and D=0? I am not very good at this it is amazing that I have this far

6. this might be helpful, 3 methods

Partial fraction - Wikipedia, the free encyclopedia

If one wants to decompose x/(x + a), then one can follow these steps:
Write as
where : and are constants
Multiply both sides by x + a
means A and B must simultaneously solve:
because the coefficients of both x and 1 have to agree, giving rise to the two equations.
Therefore
Finally, the decomposed form is:
 Distinct first-degree factors in the denominator

Suppose it is desired to decompose the rational function
into partial fractions. The denominator factors as
and so we seek scalars A and B such that
One way of finding A and B begins by "clearing fractions", i.e., multiplying both sides by the common denominator (x − 8)(x + 5). This yields
Collecting like terms gives
Equating coefficients of like terms then yields:
The solution is A = 11/13, B = 2/13. Thus we have the partial fraction decomposition
Alternatively, take the original equation
multiply by (x − 8) to get
evaluate at x = 8 to solve for A as
multiply the original equation by (x + 5) to get
evaluate at x = −5 to solve for B as
 An irreducible quadratic factor in the denominator

In order to decompose
into partial fractions, first observe that
The fact that x2 + 2x + 4 cannot be factored using real numbers can be seen by observing that the discriminant 22 − 4(1)(4) is negative. Thus we seek scalars A, B, C such that
When we clear fractions, we get
We could proceed as in the previous example, getting three linear equations in three variables A, B, and C. However, since solving such systems becomes onerous as the number of variables grows, we try a different method. Substitution of 2 for x in the identity above makes the entire second term vanish, and we get
i.e., 84 = 12A, so A = 7, and we have
Next, substitution of 0 for x yields
and so C = 4. We now have
Substitution of 1 for x yields
and so B = 3. Our partial fraction decomposition is therefore:
 A repeated first-degree factor in the denominator

Consider the rational function
The denominator factors thus:
The multiplicity of the first-degree factor (x − 4) is more than 1. In such cases, the partial fraction decomposition takes the following form:
By solving, A = −70, B = 80, C = −63. Therefore, the solution becomes
 Repeated factors in the denominator generally

For rational functions of the form
(where the p(x) may be any polynomial of sufficiently small degree) the partial fraction decomposition looks like this:
The general pattern may be quickly guessed. For rational functions of the form
with the irreducible quadratic factor x2 + 1 in the denominator (where again, the p(x) may be any polynomial of sufficiently small degree), the partial fraction decomposition looks like this:
and a similar pattern holds for any other irreducible quadratic factor.

Parāvartya Sūtra
Separation of a fractional algebraic expression into partial fractions is the reverse of the process of combining fractions by converting each fraction to the lowest common denominator (LCM) and adding the numerators. This separation is accomplished by a mental, one-line Vedic formula called the Parāvartya Sūtra[1]. Case one has fractional expressions where factors in the denominator are unique. Case two has fractional expressions where some factors may repeat as powers of a binomial.
In integral calculus we would want to write a fractional algebraic expression as the sum of its partial fractions in order to take the integral of each simple fraction separately. Once the original denominator, D0, has been factored we set up a fraction for each factor in the denominator. We may use a subscripted D to represent the denominator of the respective partial fractions which are the factors in D0. Letters A, B, C, D, E, and so on will represent the numerators of the respective partial fractions.
We calculate each respective numerator by (1) calculating the Parāvartya value of the denominator (which is the value of the variable making that binomial factor equal to zero) and (2) then substituting this value into the original expression but ignoring that factor in the denominator. Each Parāvartya value for the variable is the value which would give an undefined value to the expression since we do not divide by zero.
General formula:
Here, a, b, c, , m, and n are given integer values.
Where x = a and
and where x = b and
and where x = c and
[2]  Case one

Factorize the expression in the denominator. Set up a partial fraction for each factor in the denominator. Apply the Parāvartya Sūtra to solve for the new numerator of each partial fraction.
 Example

Set up a partial fraction for each factor in the denominator. With this framework we apply the Sūtra to solve for A, B, and C by mental math.
1. D1 is x + 1; set it equal to zero. This gives the Parāvartya value for A when x = −1.
2. Next, substitute this value of x into the fractional expression, but without D1.
3. Put this value down as the value of A.
Proceed similarly for B and C.
D2 is x + 2; For Parāvartya B use x = −2.
D3 is x + 3; For Parāvartya C use x = −3.
Thus, to solve for A, use x = −1 in the expression but without D1:
Thus, to solve for B, use x = −2 in the expression but without D2:
Thus, to solve for C, use x = −3 in the expression but without D3:
Thus,
[3]  Case two

When factors of the denominator include powers of one expression we (1) Set up a partial fraction for each unique factor and each lower power of D; (2) We set up an equation showing the relation of the numerators if all were converted to the LCD. From the equation of numerators we solve for each numerator, A, B, C, D, and so on. This equation of the numerators is an absolute identity, true for all values of x. So, we may select any value of x and solve for the numerator.[4]
 Example

Here, we set up a partial fraction for each descending power of the denominator. Then we solve for the numerators, A and B. As the Parāvartya value for A and B will be the same, x = ½, we need an additional relation in order to solve for both. To write the relation of numerators the second fraction needs another factor of (1-2x) to convert it to the LCD, giving us 3x + 5 = A + B(1 − 2x).
To solve for A: Set the denominator of the first fraction to zero, 1 − 2x = 0. Solving for x gives the Parāvartya value for A, when x = ½. When we substitute this value, x = ½, into the relation of numerators we have 3(1/2) + 5 = A + B(0). Solving for A gives us A = 3/2 + 5 = 13/2. Hence, numerator A equals six and one-half.[5]
To solve for B: Since the equation of the numerators, here, 3x + 5 = A + B(1 − 2x), is true for all values of x, pick a value for x and use it to solve for B. As we have solved for the value of A above, A = 13/2, we may use that value to solve for B.
We may pick x = 0, use A = 13/2, and then solve for B.
3x + 5 = A + B(1 − 2x) 0 + 5 = 13/2 + B(1 + 0) 10/2 = 13/2 + B −3/2 = B We may pick x = 1. Then solve for B:
3x + 5 = A + B(1 − 2x) 3 + 5 = 13/2 + B(1 − 2) 8 = 13/2 + B(−1) 16/2 = 13/2 − B B = −3/2 We may pick x = −1. Solve for B:
3x + 5 = A + B(1 − 2x) −3 + 5 = 13/2 + B(1 + 2) 4/2 = 13/2 + 3B −9/2 = 3B −3/2 = B Hence,
or
 Technique three

A third technique is an analysis of the expanded relation of the numerators. Just match up the x-terms of each degree. Then one can see the coefficients of the matching terms and solve for the missing numerator.
 Example

Converting each fraction to the LCD we have the relation in the numerators: 3x + 5 = A + B(1 − 2x). As A and B are constants, we can expand and match up the constant terms and the x-terms. The values of A and B will then be apparent.
3x + 5 = A + B(1 − 2x) 3x + 5 = A + B − 2Bx Hence, the constant terms are set equal and the x-terms are set equal:
5 = A + B and 3x = −2Bx Therefore, by setting the coefficients equal, we may solve for B:
3 = −2B −3/2 = B And then solve for A:
5 = A − 3/2 5 + 3/2 = A 13/2 = A Hence,