Evaluate the Integral

Problem

(∫_^)((7*x5+7*x2+14)/(x3−x)*d(x))

Solution

1. Perform polynomial long division because the degree of the numerator is greater than the degree of the denominator.

7*x5+7*x2+14=(7*x2+7)*(x3−x)+7*x3+7*x2−7*x+14

7*x3+7*x2−7*x+14=7*(x3−x)+7*x2+14

(7*x5+7*x2+14)/(x3−x)=7*x2+7+(7*x2+14)/(x3−x)

2. Factor the denominator of the remaining rational expression.

x3−x=x*(x2−1)=x*(x−1)*(x+1)

3. Apply partial fraction decomposition to the remainder term.

(7*x2+14)/(x*(x−1)*(x+1))=A/x+B/(x−1)+C/(x+1)

7*x2+14=A*(x−1)*(x+1)+B*x*(x+1)+C*x*(x−1)

4. Solve for the constants by substituting values for x

Let *x=0⇒14=A*(−1)⇒A=−14

Let *x=1⇒21=B(2)⇒B=21/2

Let *x=−1⇒21=C(2)⇒C=21/2

5. Rewrite the integral using the results from division and partial fractions.

(∫_^)((7*x2+7−14/x+21/(2*(x−1))+21/(2*(x+1)))*d(x))

6. Integrate term by term using the power rule and the natural logarithm rule.

(∫_^)(7*x2*d(x))=(7*x3)/3

(∫_^)(7*d(x))=7*x

(∫_^)(−14/x*d(x))=−14*ln(x)

(∫_^)(21/(2*(x−1))*d(x))=21/2*ln(x−1)

(∫_^)(21/(2*(x+1))*d(x))=21/2*ln(x+1)

7. Combine the logarithmic terms using properties of logarithms.

21/2*(ln(x−1)+ln(x+1))=21/2*ln(x2−1)

Final Answer

(∫_^)((7*x5+7*x2+14)/(x3−x)*d(x))=(7*x3)/3+7*x−14*ln(x)+21/2*ln(x2−1)+C

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