The value of _ 0 ^ 1 x^ 4 +1 x^ 2 +1 d x is,
Mathematics · JEE Main · NTA Exams — Integral Calculus
The value of \(\int_{0}^{1} \frac{x^{4}+1}{x^{2}+1} d x\) is,
- \(\frac{1}{6}(3-4 \pi)\)
- \(\frac{1}{6}(3 \pi+4)\)
- \(\frac{1}{6}(3+4 \pi)\)
- \(\frac{1}{6}(3 \pi-4)\)
Answer
(D) 1 6 (3 -4)
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