COMEDK · Maths · 28. Indefinite Integration
\(\int \dfrac{x}{x^4-16} d x=\)
- A \(\dfrac{1}{16} \log \left|\dfrac{x^2+4}{x^2-4}\right|+C\)
- B \(\dfrac{1}{16} \log \left|\dfrac{x^2-4}{x^2+4}\right|+C\)
- C \(\dfrac{1}{8} \log \left|\dfrac{x^2-4}{x^2+4}\right|+C\)
- D \(\dfrac{1}{4} \log \left|\dfrac{x^2+4}{x^2-4}\right|+C\)
Answer & Solution
Correct Answer
(B) \(\dfrac{1}{16} \log \left|\dfrac{x^2-4}{x^2+4}\right|+C\)
Step-by-step Solution
Detailed explanation
Let \(I = \int \dfrac{x}{x^4 - 16} dx\). Substitute \(u = x^2\), then \(du = 2x dx\), which implies \(x dx = \dfrac{du}{2}\). The integral becomes \(I = \int \dfrac{1}{2(u^2 - 16)} du\). Using the partial fraction decomposition…
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