CUET · MATHS · PYQ PAPER 2025
\(\int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} d x\) is equal to
- A \(\frac{1}{2} \log _e 2-1\)
- B \(2 \log _e 2\)
- C \(2 \log _e 2-2\)
- D \(4 \log _e 2\)
Answer & Solution
Correct Answer
(B) \(2 \log _e 2\)
Step-by-step Solution
Detailed explanation
\( \int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} d x = \int_{-1}^1 \left( \frac{x^3}{(|x|+1)^2} + \frac{|x|+1}{(|x|+1)^2} \right) d x \) \( = \int_{-1}^1 \frac{x^3}{(|x|+1)^2} d x + \int_{-1}^1 \frac{1}{|x|+1} d x \) \( = 0 + 2 \int_0^1 \frac{1}{x+1} d x \)…
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