\(\int_{-1}^{1}\displaystyle\frac{x^3 + |x| + 1}{x^2+2|x|+1} dx\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\log_e2\)
Since \(x^2 = |x|^2\), the denominator is
\(x^2 + 2|x| + 1 = (|x| + 1)^2\)
which is an even function. So the integrand splits into an odd part and an even part:
\( \displaystyle \frac{x^3 + |x| + 1}{(|x| + 1)^2} = \frac{x^3}{(|x| + 1)^2} + \frac{1}{|x| + 1} \)
The first part is an odd function, so its integral over [−1, 1] is 0. The second part is an even function, so its integral is twice the integral over [0, 1], where |x| = x:
\( \displaystyle I = 0 + 2\int_{0}^{1} \frac{1}{x + 1}\,dx \)
\( \displaystyle = 2\Big[\log_e(x + 1)\Big]_{0}^{1} \)
\(= 2(\log_e 2 - \log_e 1) = 2\log_e 2\)