Question

part frac 1 lecture p (2 points) evaluate the indefinite integral. \\( \int \frac{8x^2 - 53x - 69}{(x - 6)(x^2 - 3x - 18)} \\, dx = \\) preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining.

Explanation:

Step1: Factor the denominator

First, factor the quadratic in the denominator: \(x^2 - 3x - 18=(x - 6)(x + 3)\). So the denominator becomes \((x - 6)(x - 6)(x + 3)=(x - 6)^2(x + 3)\).
The integral is now \(\int\frac{8x^2 - 53x - 69}{(x - 6)^2(x + 3)}dx\).

Step2: Partial fraction decomposition

Assume \(\frac{8x^2 - 53x - 69}{(x - 6)^2(x + 3)}=\frac{A}{x - 6}+\frac{B}{(x - 6)^2}+\frac{C}{x + 3}\).
Multiply both sides by \((x - 6)^2(x + 3)\) to get:
\(8x^2 - 53x - 69 = A(x - 6)(x + 3)+B(x + 3)+C(x - 6)^2\).

Step3: Find coefficients A, B, C

• Let \(x = 6\):

\(8(6)^2 - 53(6)-69 = A(0)+B(6 + 3)+C(0)\)
\(288 - 318 - 69 = 9B\)
\(-99 = 9B\), so \(B=-11\).

• Let \(x=-3\):

\(8(-3)^2 - 53(-3)-69 = A(0)+B(0)+C(-3 - 6)^2\)
\(72 + 159 - 69 = 81C\)
\(162 = 81C\), so \(C = 2\).

• Expand the right - hand side:

\(A(x - 6)(x + 3)+B(x + 3)+C(x - 6)^2=A(x^2 - 3x - 18)+Bx + 3B+C(x^2 - 12x + 36)\)
\(=(A + C)x^2+(-3A + B-12C)x+(-18A + 3B + 36C)\)
Compare the coefficients of \(x^2\): \(A + C=8\). Since \(C = 2\), then \(A=8 - 2 = 6\).

So the partial fraction decomposition is \(\frac{6}{x - 6}-\frac{11}{(x - 6)^2}+\frac{2}{x + 3}\).

Step4: Integrate term by term

Integrate each term separately:

• \(\int\frac{6}{x - 6}dx=6\ln|x - 6|+C_1\)
• \(\int\frac{-11}{(x - 6)^2}dx=-11\int(x - 6)^{-2}dx\). Using the power rule \(\int u^n du=\frac{u^{n + 1}}{n+1}+C\) (\(n

eq - 1\)), let \(u=x - 6\), \(n=-2\), we get \(-11\times\frac{(x - 6)^{-1}}{-1}+C_2=\frac{11}{x - 6}+C_2\)

• \(\int\frac{2}{x + 3}dx=2\ln|x + 3|+C_3\)

Answer:

\(6\ln|x - 6|+\frac{11}{x - 6}+2\ln|x + 3|+C\) (where \(C = C_1 + C_2+C_3\) is the constant of integration)