JAMB 2013 Mathematics · Question 9

Integrate 1+xx3dx\frac{1 + x}{x^{3}} \mathrm d x

Options

  1. A.2x2−1x+k2x^{2} - \frac{1}{x} + k
  2. B.−12x2−1x+k-\frac{1}{2x^{2}} - \frac{1}{x} + k
  3. C.−x22−1x+k-\frac{x^{2}}{2} - \frac{1}{x} + k
  4. D.x2−1x+kx^{2} - \frac{1}{x} + k

Answer: B. −12x2−1x+k-\frac{1}{2x^{2}} - \frac{1}{x} + k

Explanation written with AI and checked against the answer key

Analysis

To solve this problem, we need to integrate the function ∫1+xx3 dx\int \frac{1 + x}{x^3} \, dx. The key strategy here is to simplify the integrand by splitting the fraction before applying the power rule for integration.

Concept: The Power Rule for Integration

The power rule states that:

∫xn dx=xn+1n+1+C(where n≠−1)\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (\text{where } n \neq -1)
Step-by-Step Solution

Step 1: Split the fraction Divide each term in the numerator by the denominator (x3x^3):

∫(1x3+xx3) dx\int \left( \frac{1}{x^3} + \frac{x}{x^3} \right) \, dx

Step 2: Simplify using indices Rewrite the terms using negative exponents:

∫(x−3+x−2) dx\int (x^{-3} + x^{-2}) \, dx

Step 3: Integrate term by term Apply the power rule ∫xn dx=xn+1n+1\int x^n \, dx = \frac{x^{n+1}}{n+1} to each term:

  • For x−3x^{-3}: $∫x−3 dx=x−3+1−3+1=x−2−2=−12x2\int x^{-3} \, dx = \frac{x^{-3+1}}{-3+1} = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$
  • For x−2x^{-2}: $∫x−2 dx=x−2+1−2+1=x−1−1=−1x\int x^{-2} \, dx = \frac{x^{-2+1}}{-2+1} = \frac{x^{-1}}{-1} = -\frac{1}{x}$

Step 4: Combine the results Adding the constant of integration (kk):

−12x2−1x+k-\frac{1}{2x^2} - \frac{1}{x} + k
Conclusion

Comparing our result with the given options, we find that it matches option B.

Exam Tip

When you see a fraction with a single term in the denominator (like x3x^3), always try to split it into separate fractions first. It makes the integration much simpler and prevents errors associated with the quotient rule (which doesn't apply to integration).

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