The value of int_{e^2}^{e^4} frac{1}{x} left( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $I = \int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{((\ln x)^2+1)^{-1}}}{e^{((\ln x)^2+1)^{-1}} + e^{((6-\ln x)^2+1)^{-1}}} \right) dx$. Substit

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Let $I = \int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{((\ln x)^2+1)^{-1}}}{e^{((\ln x)^2+1)^{-1}} + e^{((6-\ln x)^2+1)^{-1}}} \right) dx$. Substitute $\ln x = t$,then $\frac{1}{x} dx = dt$. When $x = e^2$,$t = 2$. When $x = e^4$,$t = 4$. So,$I = \int_{2}^{4} \frac{e^{(t^2+1)^{-1}}}{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}} dt$. Using the property $\int_{a}^{b} f(t) dt = \int_{a}^{b} f(a+b-t) dt$,we get: $I = \int_{2}^{4} \frac{e^{((6-t)^2+1)^{-1}}}{e^{((6-t)^2+1)^{-1}} + e^{(t^2+1)^{-1}}} dt$. Adding the two expressions for $I$: $2I = \int_{2}^{4} \frac{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}}{e^{(t^2+1)^{-1}} + e^{((6-t)^2+1)^{-1}}} dt = \int_{2}^{4} 1 dt = [t]_{2}^{4} = 4 - 2 = 2$. Therefore,$I = 1$.

The value of $\int_{e^2}^{e^4} \frac{1}{x} \left( \frac{e^{((\ln x)^2+1)^{-1}}}{e^{((\ln x)^2+1)^{-1}} + e^{((6-\ln x)^2+1)^{-1}}} \right) dx$ is

Similar Questions

$\int_3^5(x-3)^3(5-x)^5 d x=$

$\int_0^{\pi /2} \frac{\cos x}{1 + \cos x + \sin x} \,dx = $

$\int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x=$

The value of the definite integral $\int_{2}^{4} (x(3 - x)(4 + x)(6 - x)(10 - x) + \sin x) dx$ equals

Evaluate: $\int_{-a}^a f(x) dx - \int_0^a f(-x) dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module