JEE Main 2026 — Mathematics Calculus

Let $\displaystyle I = \int_{-1}^{1} \left(\dfrac{x^3 + |x| + 1}{x^2 + 2|x| + 1}\right) dx$I=∫−11​(x2+2∣x∣+1x3+∣x∣+1​)dx

We can split the integral into two parts:

$$I = \int_{-1}^{1} \dfrac{x^3}{x^2 + 2|x| + 1} dx + \int_{-1}^{1} \dfrac{|x| + 1}{x^2 + 2|x| + 1} dx$$I=∫−11​x2+2∣x∣+1x3​dx+∫−11​x2+2∣x∣+1∣x∣+1​dx

The first integrand $f(x) = \dfrac{x^3}{x^2 + 2|x| + 1}$f(x)=x2+2∣x∣+1x3​ is an odd function since $f(-x) = -f(x)$f(−x)=−f(x). Therefore, its integral over the symmetric interval $[-1, 1]$[−1,1] is zero.

The second integrand $g(x) = \dfrac{|x| + 1}{x^2 + 2|x| + 1}$g(x)=x2+2∣x∣+1∣x∣+1​ is an even function since $g(-x) = g(x)$g(−x)=g(x). Therefore, its integral over $[-1, 1]$[−1,1] is twice the integral over $[0, 1]$[0,1].

$$I = 0 + 2 \int_{0}^{1} \dfrac{x + 1}{x^2 + 2x + 1} dx$$I=0+2∫01​x2+2x+1x+1​dx

$$I = 2 \int_{0}^{1} \dfrac{x + 1}{(x + 1)^2} dx$$I=2∫01​(x+1)2x+1​dx

$$I = 2 \int_{0}^{1} \dfrac{1}{x + 1} dx$$I=2∫01​x+11​dx

$$I = 2 [\log_e(x + 1)]_{0}^{1}$$I=2[loge​(x+1)]01​

$$I = 2(\log_e 2 - \log_e 1) = 2\log_e 2$$I=2(loge​2−loge​1)=2loge​2

Answer: $2\log_e 2$2loge​2