Question

Resolve this into partial fraction:x^2+x+4/(1-x)(x^2+1)

Answer 1

To resolve \( \frac{x^2 + x + 4}{(1 - x)(x^2 + 1)} \) into partial fractions, we first express the fraction in the form:

\[
\frac{x^2 + x + 4}{(1 - x)(x^2 + 1)} = \frac{A}{1 - x} + \frac{Bx + C}{x^2 + 1}
\]

where \( A \), \( B \), and \( C \) are constants we need to determine.

Next, we multiply both sides by the denominator \( (1 - x)(x^2 + 1) \) to eliminate the denominators:

\[
x^2 + x + 4 = A(x^2 + 1) + (Bx + C)(1 - x)
\]

Expanding the right side:

\[
x^2 + x + 4 = A(x^2 + 1) + (Bx + C)(1 - x)
\]
\[
= A(x^2) + A(1) + (Bx + C)(1 - x)
\]
\[
= Ax^2 + A + Bx - Bx^2 + C - Cx
\]
\[
= (A - B)x^2 + (B - C)x + (A + C)
\]

Now we can equate the coefficients of corresponding terms from both sides:

1. Coefficient of \( x^2 \): \( A - B = 1 \)
2. Coefficient of \( x \): \( B - C = 1 \)
3. Constant term: \( A + C = 4 \)

We now have a system of linear equations:

1. \( A - B = 1 \) (Equation 1)
2. \( B - C = 1 \) (Equation 2)
3. \( A + C = 4 \) (Equation 3)

Now, we can start solving these equations. From Equation 1, we have:

\[
A = B + 1
\]

Substituting \( A \) into Equation 3:

\[
(B + 1) + C = 4
\]
\[
B + C + 1 = 4
\]
\[
B + C = 3 \quad \text{(Equation 4)}
\]

Now we can use Equation 4 and Equation 2 together. From Equation 2, we can express \( C \):

\[
C = B - 1
\]

Substituting \( C \) into Equation 4:

\[
B + (B - 1) = 3
\]
\[
2B - 1 = 3
\]
\[
2B = 4
\]
\[
B = 2
\]

Now substituting \( B = 2 \) back to find \( C \):

\[
C = B - 1 = 2 - 1 = 1
\]

And substituting \( B = 2 \) to find \( A \):

\[
A = B + 1 = 2 + 1 = 3
\]

Now we have:

\[
A = 3, \quad B = 2, \quad C = 1
\]

Thus, the partial fraction decomposition of \( \frac{x^2 + x + 4}{(1 - x)(x^2 + 1)} \) is:

\[
\frac{3}{1 - x} + \frac{2x + 1}{x^2 + 1}
\]

So we can write the final result as:

\[
\frac{x^2 + x + 4}{(1 - x)(x^2 + 1)} = \frac{3}{1 - x} + \frac{2x + 1}{x^2 + 1}
\]