12. Techniques of Integration

Partial Fractions

12. Techniques of Integration

Partial Fractions

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Multiple Choice

Express the integrand as a sum of partial fractions and evaluate the integral.
$\int_{}^{} \frac{2}{x^{3} + x^{2} - x - 1} d x$

$\int_{}^{} (- \frac{1}{2 (x + 1)} - \frac{1}{{(x + 1)}^{2}} + \frac{1}{2 (x - 1)}) d x = - \frac{1}{2} \ln ∣ x + 1 ∣ + \ln ∣ (x + 1)^{2} ∣ + \frac{1}{2} \ln ∣ x - 1 ∣ + C$

$\int_{}^{} (- \frac{2}{(x + 1)} - \frac{2}{{(x + 1)}^{2}} + \frac{2}{(x - 1)}) d x = - 2 \ln ∣ x + 1 ∣ + \frac{2}{x + 1} + 2 \ln ∣ x - 1 ∣ + C$

$\int_{}^{} (- \frac{1}{2 (x + 1)} - \frac{1}{{(x + 1)}^{2}} + \frac{1}{2 (x - 1)}) d x = - \frac{1}{2} \ln ∣ x + 1 ∣ + \frac{1}{x + 1} + \frac{1}{2} \ln ∣ x - 1 ∣ + C$

$\int_{}^{} (- \frac{2}{(x + 1)} - \frac{2}{{(x + 1)}^{2}} + \frac{2}{(x - 1)}) d x = - 2 \ln ∣ x + 1 ∣ + 2 \ln ∣ {(x + 1)}^{2} ∣ + 2 \ln ∣ x - 1 ∣ + C$

Verified step by step guidance

Step 1: Begin by factoring the denominator of the integrand. The denominator is $x^3 + x^2 - x - 1$. Group terms to factorize: $(x^3 + x^2) - (x + 1) = x^2(x + 1) - 1(x + 1) = (x + 1)(x^2 - 1)$. Further factorize $x^2 - 1$ as $(x - 1)(x + 1)$. Thus, the denominator becomes $(x + 1)^2(x - 1)$.

Step 2: Rewrite the integrand $\frac{2}{x^3 + x^2 - x - 1}$ as $\frac{2}{(x + 1)^2(x - 1)}$. Decompose this into partial fractions: $\frac{2}{(x + 1)^2(x - 1)} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x - 1}$, where $A, B,$ and $C$ are constants to be determined.

Step 3: Multiply through by the denominator $(x + 1)^2(x - 1)$ to eliminate the fractions: $2 = A(x + 1)(x - 1) + B(x - 1) + C(x + 1)^2$. Expand and collect like terms to solve for $A, B,$ and $C$.

Step 4: Solve for the constants $A, B,$ and $C$ by substituting convenient values for $x$ (e.g., $x = -1$, $x = 1$) and equating coefficients of powers of $x$. After solving, you will find $A = -\frac{1}{2}, B = -1,$ and $C = \frac{1}{2}$.

Step 5: Substitute the partial fraction decomposition back into the integral: $\int \frac{2}{(x + 1)^2(x - 1)} dx = \int \left(-\frac{1}{2(x + 1)} - \frac{1}{(x + 1)^2} + \frac{1}{2(x - 1)}\right) dx$. Integrate each term separately to obtain the result: $-\frac{1}{2}\ln|x + 1| - \frac{1}{x + 1} + \frac{1}{2}\ln|x - 1| + C$.