The value of the integral int frac{d x}{left( | vedclass

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

Question

(C) Let $I = \int \frac{dx}{(e^x + e^{-x})^2}$.Multiply the numerator and denominator by $e^{2x}$:$I = \int \frac{e^{2x} dx}{(e^x \cdot e^x + e^{-x} \

Vedclass · Accepted Answer

$-\frac{1}{2}\left(e^{2 x}+1\right)^{-1}+C$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(e^x + e^{-x})^2}$.Multiply the numerator and denominator by $e^{2x}$:$I = \int \frac{e^{2x} dx}{(e^x \cdot e^x + e^{-x} \cdot e^x)^2} = \int \frac{e^{2x} dx}{(e^{2x} + 1)^2}$.Let $u = e^{2x} + 1$. Then $du = 2e^{2x} dx$,which implies $e^{2x} dx = \frac{du}{2}$.Substituting these into the integral:$I = \int \frac{du/2}{u^2} = \frac{1}{2} \int u^{-2} du$.Integrating with respect to $u$:$I = \frac{1}{2} \left( \frac{u^{-1}}{-1} \right) + C = -\frac{1}{2u} + C$.Substituting $u = e^{2x} + 1$ back:$I = -\frac{1}{2(e^{2x} + 1)} + C = -\frac{1}{2}(e^{2x} + 1)^{-1} + C$.

The value of the integral $\int \frac{d x}{\left(e^x+e^{-x}\right)^2}$ is

Similar Questions

$\int {\frac{{1 - {x^7}}}{{x(1 + {x^7})}}} \,dx$ equals

$\int {\frac{{{e^{2x}} + 1}}{{{e^{2x}} - 1}}} \,dx$ equals

Integrate the function $\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}$.

$\int \sin^5 x \cos^4 x \, dx = $

$\int \frac{1}{\cos x \sqrt{\cos 2 x}} \, dx = $ (where $C$ is a constant of integration.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module