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

The value of $\int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \,dx$ is

If $I = \int_0^\pi x \left\{ \sin^2(\sin x) + \cos^2(\cos x) \right\} dx$,then $[I] = \ldots$. Here,$[.]$ denotes the greatest integer function.

$\int_0^{\pi /2} {x\cot x\,dx} $ equals

$\int_{ - \pi /2}^{\pi /2} {\log \left( {\frac{{2 - \sin \theta }}{{2 + \sin \theta }}} \right)\,d\theta = } $

Let the function $F$ be defined as $F(x) = \int_{1}^{x} \frac{e^{t}}{t} dt, x > 0$. Then the value of the integral $\int_{1}^{x} \frac{e^{t}}{t+a} dt$,where $a > 0$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module