Question

integration of e^x /√1-e^x with respect to dx​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\int\dfrac{{e}^{x}}{\sqrt{1-{e}^{x}}}\,dx$

$\bf{Put\,\,\,\,1-{e}^{x}={t}^{2}}$

$\bf{\mapsto\,\,\,-{e}^{x}\,dx=2{t}\,dt}$

$\bf{\mapsto\,\,\,{e}^{x}\,dx=-2{t}\,dt}$

$\displaystyle=\int\dfrac{-2t}{\sqrt{{t}^{2}}}\,dt$

$\displaystyle=-2\int\dfrac{t\,dt}{t}$

$\displaystyle=-2\int\,dt$

$\displaystyle=-2\,t+C$

$\displaystyle=-2\sqrt{1-{e}^{x}}+C$

Answer 2

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int \dfrac{ {e}^{x} }{ \sqrt{1 - {e}^{x} } } \: dx$

Let us assume that:

$\rm \longrightarrow u =1 - {e}^{x}$

$\rm \longrightarrow \dfrac{du}{dx} = - {e}^{x}$

$\rm \longrightarrow du= -{e}^{x} \: dx$

Therefore, the integral changes to:

$\displaystyle \rm \longrightarrow I = \int \dfrac{ - du}{ \sqrt{u}}$

$\displaystyle \rm \longrightarrow I = - \int \dfrac{du}{ \sqrt{u}}$

$\displaystyle \rm \longrightarrow I = - \dfrac{ {u}^{ \frac{ - 1}{2} + 1} }{ \frac{ - 1}{2} + 1} + C$

$\displaystyle \rm \longrightarrow I = - \dfrac{ \sqrt{u} }{ \frac{1}{2}} + C$

$\displaystyle \rm \longrightarrow I = -2 \sqrt{u} + C$

Substituting back the value of u, we get:

$\displaystyle \rm \longrightarrow I = -2 \sqrt{1 -{e}^{x} } + C$

Therefore:

$\displaystyle \rm \longrightarrow \int \dfrac{ {e}^{x} }{ \sqrt{1 - {e}^{x} } } \: dx = -2 \sqrt{1 -{e}^{x} } + C$

★ Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$