int frac{dx}{e^x + 1 - 2e^{-x}} = | vedclass

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

Question

(C) Let $I = \int \frac{dx}{e^x + 1 - 2e^{-x}}$.Multiply numerator and denominator by $e^x$:$I = \int \frac{e^x dx}{e^{2x} + e^x - 2}$.Let $e^x = t$,t

Vedclass · Accepted Answer

$\frac{1}{3}\log (e^x - 1) - \frac{1}{3}\log (e^x + 2) + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{e^x + 1 - 2e^{-x}}$.Multiply numerator and denominator by $e^x$:$I = \int \frac{e^x dx}{e^{2x} + e^x - 2}$.Let $e^x = t$,then $e^x dx = dt$.$I = \int \frac{dt}{t^2 + t - 2} = \int \frac{dt}{(t + 2)(t - 1)}$.Using partial fractions:$\frac{1}{(t + 2)(t - 1)} = \frac{1}{3} \left[ \frac{1}{t - 1} - \frac{1}{t + 2} \right]$.Integrating both sides:$I = \frac{1}{3} \int \left( \frac{1}{t - 1} - \frac{1}{t + 2} \right) dt = \frac{1}{3} \log |t - 1| - \frac{1}{3} \log |t + 2| + c$.Substituting $t = e^x$ back:$I = \frac{1}{3} \log (e^x - 1) - \frac{1}{3} \log (e^x + 2) + c$.

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

Similar Questions

Integrate the rational function: $\frac{3x+5}{x^{3}-x^{2}-x+1}$

$\int \frac{d x}{(\sin x+\cos x)(2 \cos x+\sin x)} = $

$\int_{\log 4}^{\log 5} \frac{e^{2 x}+e^x}{e^{2 x}-5 e^x+6} d x=$

Integrate the rational function: $\frac{1}{x^{4}-1}$

$\int \frac{d x}{x\left(x^2+1\right)}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module