Question

Integrate sqrt(e^x-1) dx​

Answer 1

Solution:-

We need to evaluate the integral,

$\displaystyle I = \int \sqrt{ {e}^{x} - 1 } dx$

Let's substitute $t = \sqrt{e^x-1}$

$\implies \sqrt{ {e}^{x} - 1 } = t$

Squaring both sides

$\implies {e}^{x} - 1 = {t}^{2} ...(1.)$

Differentiation both sides

$\implies {e}^{x} dx = 2 t \: dt$

$\implies dx = \dfrac{2 t \: dt }{e^ {x}}$

${\implies dx = \dfrac{2 t \: dt }{ {t}^{2} + 1} \qquad \sf\left(using \: eq.1 \right)}$

Now, we have :

$\displaystyle I = \int \sqrt{ {e}^{x} - 1 } dx$

Substitute for t and dx

$\implies \displaystyle I = \int t. \frac{2t}{ {t}^{2} + 1} dt$

$\implies \displaystyle I = 2 \int \frac{ {t}^{2} }{ {t}^{2} + 1} dt$

$\implies \displaystyle I = 2 \int \frac{ {t}^{2} + 1 - 1}{ {t}^{2} + 1} dt$

${ \implies \displaystyle I = 2 \left \{ \int \frac{ {t}^{2} + 1 }{ {t}^{2} + 1} dt - \int \frac{1}{ {t}^{2} + 1} dt\right \}}$

${ \implies \displaystyle I = 2 \left \{ \int dt - \tan^{ - 1} t+ C\right \}}$

${ \implies \displaystyle I = 2 \left \{ t - \tan^{ - 1} t+ C\right \}}$

Substitute back value of t

${ \implies \displaystyle I = 2 \left \{ \sqrt{ {e}^{x} - 1 } - \tan^{ - 1} ( \sqrt{ {e}^{x} - 1 } )+ c\right \}}$

Therefore the required answer is,

$\underline{\boxed{ \displaystyle \int \sqrt{ {e}^{x} - 1 } dx = 2 \left ( \sqrt{ {e}^{x} - 1 } - \tan^{ - 1} ( \sqrt{ {e}^{x} - 1 } )+ c\right )}}$

Formula used,

• $\boxed{\int dx = x + C}$
• $\boxed{\int \dfrac{1}{x^2+1} dx= \tan^{-1}x + C}$
• $\boxed{\frac{d}{dx} e^x = e^x}$
• $\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$