int frac{d x}{left(e^x+e^{-x}right)^2}= | vedclass

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

Question

(B) माना $I = \int \frac{d x}{\left(e^x+e^{-x}\right)^2} = \int \frac{d x}{\left(e^x+\frac{1}{e^x}\right)^2}$ $= \int \frac{e^{2 x} d x}{\left(e^{2 x}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) माना $I = \int \frac{d x}{\left(e^x+e^{-x}\right)^2} = \int \frac{d x}{\left(e^x+\frac{1}{e^x}\right)^2}$ $= \int \frac{e^{2 x} d x}{\left(e^{2 x}+1\right)^2}$ माना $t = e^{2 x}+1$, तब $dt = 2e^{2 x} dx$, जिसका अर्थ है कि $e^{2 x} dx = \frac{dt}{2}$. इन मानों को समाकलन में प्रतिस्थापित करने पर: $I = \int \frac{1}{2} \frac{dt}{t^2} = \frac{1}{2} \int t^{-2} dt$ $= \frac{1}{2} \left( \frac{t^{-1}}{-1} \right) + c = -\frac{1}{2t} + c$ $t = e^{2 x}+1$ वापस रखने पर: $I = -\frac{1}{2\left(e^{2 x}+1\right)} + c$

$\int \frac{d x}{\left(e^x+e^{-x}\right)^2}=$

Similar Questions

$\int \frac{1}{\cos^2 x (1 - \tan x)^2} dx = $

$\int \frac{d x}{4+5 \cos x} = $

यदि $\int \cos ^{\frac{3}{5}} x \cdot \sin ^3 x \,d x = \frac{-1}{m} \cos ^{m} x + \frac{1}{n} \cos ^{n} x + c$ है,(जहाँ $c$ समाकलन का स्थिरांक है),तो $(m, n) = $

$\int \frac{\tan(\log x)}{x} \, dx = $

$\int(x+1)(x+2)^4(x+3) \, dx$ का मान ज्ञात कीजिए।

Vedclass Test Series

Exam Paper Generator

Online Exam Module