WBJEE · Maths · Definite Integration
\(\int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} d x\) is equal to
- A \(\log 2\)
- B \(2 \log 2\)
- C \(\frac{1}{2} \log 2\)
- D \(4 \log 2\)
Answer & Solution
Correct Answer
(B) \(2 \log 2\)
Step-by-step Solution
Detailed explanation
\(I=\int_{-1}^1 \frac{x^3+|x|+1}{x^2+2|x|+1} d x\) Applying King's property \(\mathrm{I}=\int_{-1}^1 \frac{-\mathrm{x}^3+|\mathrm{x}|+1}{\mathrm{x}^2+2|\mathrm{x}|+1} \mathrm{dx}\) Applying \(I=\int_{-1}^1 \frac{|x|+1}{x^2+2|x|+1} d x\) being even function…
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