AP EAMCET · Maths · Definite Integration
\(\int_0^1 \frac{x^4+1}{x^6+1} d x=\)
- A \(\frac{\pi}{3}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{6}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\int_0^1 \frac{x^4+1}{x^6+1} d x = \int_0^1 \left( \frac{1}{x^2+1} + \frac{x^2}{x^6+1} \right) d x\) \(= \int_0^1 \frac{1}{x^2+1} d x + \int_0^1 \frac{x^2}{(x^3)^2+1} d x\) \(= [\arctan(x)]_0^1 + \frac{1}{3} [\arctan(x^3)]_0^1\)…
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