Question

express the integrand as a sum of partial fractions and evaluate the integral.
int\frac{9x^{2}+6x + 9}{x^{3}-1}dx
express the integrand as a sum of partial fractions.
\frac{9x^{2}+6x + 9}{x^{3}-1}=square
(simplify your answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Factor the denominator

We know that \(x^{3}-1=(x - 1)(x^{2}+x + 1)\). So we assume \(\frac{9x^{2}+6x + 9}{x^{3}-1}=\frac{A}{x - 1}+\frac{Bx+C}{x^{2}+x + 1}\).

Step2: Combine the right - hand side

\(\frac{A}{x - 1}+\frac{Bx+C}{x^{2}+x + 1}=\frac{A(x^{2}+x + 1)+(Bx + C)(x - 1)}{(x - 1)(x^{2}+x + 1)}=\frac{(A + B)x^{2}+(A - B + C)x+(A - C)}{x^{3}-1}\).

Step3: Equate the numerators

We have the system of equations \(

$$\begin{cases}A + B=9\\A - B + C=6\\A - C=9\end{cases}$$

\).
From \(A - C=9\), we get \(C=A - 9\).
Substitute \(C = A - 9\) into \(A - B + C=6\), we have \(A - B+A - 9=6\), or \(2A - B=15\).
Combined with \(A + B=9\), add the two equations: \((2A - B)+(A + B)=15 + 9\), \(3A=24\), so \(A = 8\).
Since \(A = 8\), from \(A + B=9\), we get \(B=1\).
Since \(A = 8\), from \(C=A - 9\), we get \(C=-1\).

Step4: Write the partial - fraction decomposition

\(\frac{9x^{2}+6x + 9}{x^{3}-1}=\frac{8}{x - 1}+\frac{x - 1}{x^{2}+x + 1}\)

Answer:

\(\frac{8}{x - 1}+\frac{x - 1}{x^{2}+x + 1}\)