JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Let I = _ -1 ^ 1 ( x^3 + |x| + 1 x^2 + 2|x| + 1 ) dx We can split the integral into two parts: I = _ -1 ^ 1 x^3 x^2 + 2|x| + 1 dx + _ -1 ^ 1 |x| + 1 x^2 + 2|x| + 1 dx The first integrand f(x) = x^3 x^2 + 2|x| + 1 is an odd function since f(-x) = -f(x). Therefore, its integral over the symmetric interval [-1, 1] is zero. The second integrand g(x) = |x| + 1 x^2 + 2|x| + 1 is an even function since g(-x) = g(x). Therefore, its integral over [-1, 1] is twice the integral over [0, 1]. I = 0 + 2 _ 0 ^ 1 x + 1 x^2 + 2x + 1 dx I = 2 _ 0 ^ 1 x + 1 (x + 1)^2 dx I = 2 _ 0 ^ 1 1 x + 1 dx I = 2 [ _e(x + 1)]_ 0 ^ 1 I = 2( _e 2 - _e 1) = 2 _e 2 Answer: 2 _e 2