Mathematics · 2023
JEE Main · 11 April 2023, Shift 1 · Q15
The value of the integral ∫_- log_e 2^log_e 2 e^x( log_e(e^x+√(1+e^2 x))) d x is equal to
The value of the integral $\displaystyle \int_{-\log _e 2}^{\log _e 2} e^x\left(\log _e\left(e^x+\sqrt{1+e^{2 x}}\right)\right) d x$ is equal to
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}$
More from Definite Integration
- Let the function f:[0,2] → R be defined as f(x)= {e^min {x^2, x-[x]}, x ∈[0,1); e^[x- log_e x], x ∈[1,2]} where [t] denotes the greatest integer less…2023
- The value of k ∈ N for which the integral I_n=∫_0^1(1-x^k)^n d x, n ∈ N, satisfies 147 I_20=148 I_21 is2024
- ∫_0^π / 4 (cos^2 x sin^2 x)/(( cos^3 x+ sin^3 x)^2) d x is equal to2024
- If ∫_0^1 4 cot^-1(1-2 x+4 x^2) d x=a tan^-1(2)-b log_e(5), where a, b ∈ N, then (2 a+b) is equal to ____.2026
- Let f(x) be a function satisfying f(x)+f(π-x)=π^2, ∀ x ∈ R. Then ∫_0^π f(x) sin x d x is equal to2023
- Let for f(x)=7 tan^8 x+7 tan^6 x-3 tan^4 x-3 tan^2 x, I_1=∫_0^π / 4 f(x) d x and I_2=∫_0^π / 4 x f(x) d x. Then 7 I_1+12 I_2 is equal to:2025
- The integral 80 ∫_0^((π)/4)((sin θ+ cos θ)/(9+16 sin 2 θ)) d θ is equal to:2025
- lim_n → ∞[1/(1+n)+1/(2+n)+1/(3+n)+…+1/(2 n)] is equal to2023
JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.