Value of \( \int_{-1}^{1}\displaystyle\frac{x^5+|x|+1}{x^2+2|x|+1}\,dx \) is equal to :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \log_e 4 \)
Since \( x^2=|x|^2 \), the denominator is \( x^2+2|x|+1=(|x|+1)^2 \). Splitting the integrand lets us use the properties of odd and even functions on [−a, a].
\( \displaystyle \int_{-1}^{1}\frac{x^5+|x|+1}{x^2+2|x|+1}\,dx=\int_{-1}^{1}\frac{x^5}{(|x|+1)^2}\,dx+\int_{-1}^{1}\frac{|x|+1}{(|x|+1)^2}\,dx \)
The first integrand is odd (\( x^5 \) is odd and the denominator is even), so its integral over [−1, 1] is 0. The second integrand is even:
\( \displaystyle \int_{-1}^{1}\frac{1}{|x|+1}\,dx=2\int_{0}^{1}\frac{1}{x+1}\,dx \)
\( \displaystyle =2\Big[\log_e(x+1)\Big]_{0}^{1} \)
\( \displaystyle =2\log_e 2 \)
\( \displaystyle =\log_e 4 \)
Why the other options are incorrect
Option (1) 0 — Incorrect. Only the \( x^5 \) part is odd; the part \( \frac{|x|+1}{(|x|+1)^2} \) is even and positive, so the integral is not 0.
Option (2) \( \log_e 2 \) — Incorrect. \( \log_e 2 \) is only \( \int_{0}^{1}\frac{dx}{x+1} \); for an even function the integral over [−1, 1] is twice this.
Option (4) \( \displaystyle\frac{1}{2}\log_e 2 \) — Incorrect. The value is \( 2\log_e 2 \), not \( \frac{1}{2}\log_e 2 \).