91Ó°ÊÓ

We need to factor the polynomial in the denominator to proceed with partial fractions decomposition: $$x^3-2x^2-4x+8$$ Since the polynomial does not have an obvious factor, we can use either synthetic division or the Rational Root Theorem to find a root and then factor the polynomial. In this case, we will try the Rational Root Theorem, which says that if \(\dfrac{p}{q}\) is a rational root of the polynomial, then \(p\) divides the constant term (8) and \(q\) divides the leading coefficient (1). So, candidates for rational roots are \(\pm1\), \(\pm2\), \(\pm4\), and \(\pm8\). By evaluating the polynomial at each possible root, we find that \(x=2\) is a root. So, we can factor out \((x-2)\) from the polynomial: $$x^3-2x^2-4x+8=(x-2)(x^2-4)$$ Now, the denominator can be factored further into \((x-2)(x+2)(x-2)\).

Now, we will decompose the integrand using partial fraction decomposition. Since the denominator has different linear factors, we can write: $$\frac{1}{(x-2)^2(x+2)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+2}$$ To find the constants \(A\), \(B\), and \(C\), we will clear the fractions by multiplying both sides with the denominator \((x-2)^2(x+2)\): $$1 = A(x-2)(x+2) + B(x+2) + C(x-2)^2$$

Now, we will plug in suitable values for \(x\) to solve for \(A\), \(B\), and \(C\). 1. Let \(x = 2\): $$1 = A(0) + B(4) + C(0) \Rightarrow B = \frac{1}{4}$$ 2. Let \(x = -2\): $$1 = A(0) + B(0) + C(16) \Rightarrow C = \frac{1}{16}$$ 3. Let \(x = 0\) (or any other value not used yet): $$1 = 4A + \frac{1}{2} + 4C \Rightarrow A = \frac{1}{8}$$ So, the decomposed integrand is: $$\frac{1}{8(x-2)} + \frac{1}{4(x-2)^2} + \frac{1}{16(x+2)}$$

Now, we will integrate term-by-term: $$\int\frac{1}{(x-2)^2(x+2)}~dx = \int \left(\frac{1}{8(x-2)}+\frac{1}{4(x-2)^2}+\frac{1}{16(x+2)}\right) dx$$ $$ = \frac{1}{8} \int \frac{1}{x-2}~dx + \frac{1}{4}\int \frac{1}{(x-2)^2}~dx + \frac{1}{16}\int \frac{1}{x+2}~dx$$ Now, integrate each term: $$ = \frac{1}{8} \ln|x-2| - \frac{1}{4(x-2)} + \frac{1}{16} \ln|x+2| + C$$ So, the given integral evaluates to: $$\int \frac{d x}{x^{3}-2 x^{2}-4 x+8} = \frac{1}{8} \ln|x-2| - \frac{1}{4(x-2)} + \frac{1}{16} \ln|x+2| + C$$