AP EAMCET · Maths · Indefinite Integration
\(\int \frac{x+1}{x^3-1} d x=\)
- A \(\frac{1}{3} \log \left(\frac{x+1}{x^2+x+1}\right)+c\)
- B \(\frac{1}{3} \log \left(\frac{(x-1)^2}{x^2+x+1}\right)+c\)
- C \(\frac{1}{3} \log \left(\frac{x-1}{x^2+x+1}\right)+c\)
- D \(\frac{1}{3} \log \left(\frac{(x+1)^2}{x^2-x+1}\right)+c\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3} \log \left(\frac{(x-1)^2}{x^2+x+1}\right)+c\)
Step-by-step Solution
Detailed explanation
\(\frac{x+1}{x^3-1} = \frac{x+1}{(x-1)(x^2+x+1)} = \frac{2/3}{x-1} - \frac{(2x+1)/3}{x^2+x+1}\) \(\int \left( \frac{2}{3(x-1)} - \frac{2x+1}{3(x^2+x+1)} \right) d x\) \(\frac{2}{3} \log|x-1| - \frac{1}{3} \log|x^2+x+1| + C\)…
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