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Question Number 143100 by lapache last updated on 10/Jun/21

\int \sqrt{e^{x} + 1} = \dots . . ? ? ?

Answered by qaz last updated on 10/Jun/21

\int \sqrt{e^{x} + 1} dx

= \int \sqrt{u + 1} \frac{du}{u}

= \int \frac{u + 1}{u \sqrt{u + 1}} du

= 2 \sqrt{u + 1} + \int \frac{du}{u^{2} \sqrt{\frac{1}{u} + \frac{1}{u^{2}}}}

= 2 \sqrt{u + 1} - \int \frac{d (\frac{1}{u})}{\sqrt{{(\frac{1}{u} + \frac{1}{2})}^{2} - \frac{1}{4}}}

= 2 \sqrt{u + 1} - \int \frac{d [2 (\frac{1}{u} + \frac{1}{2})]}{\sqrt{[4 {(\frac{1}{u} + \frac{1}{2})}^{2} - 1]}}

= 2 \sqrt{u + 1} - arccosh (2 (\frac{1}{u} + \frac{1}{2})) + C

= 2 \sqrt{e^{x} + 1} - arccosh (\frac{2 + e^{x}}{e^{x}}) + \overset{}{C}

Answered by MJS_new last updated on 10/Jun/21

\int \sqrt{e^{x} + 1} d x =

[t = \sqrt{e^{x} + 1} \to d x = \frac{2 \sqrt{e^{x} + 1}}{e^{x}} d t]

= 2 \int \frac{t^{2}}{t^{2} - 1} d t = \int (2 + \frac{1}{t - 1} - \frac{1}{t + 1}) d t =

= 2 t + \ln \frac{t - 1}{t + 1} = \dots

= 2 \sqrt{e^{x} + 1} - x + \ln (e^{x} + 2 - 2 \sqrt{e^{x} + 1}) + C

Answered by pticantor last updated on 10/Jun/21

w o k o o o e t t u v e u x l e r i z p o u r c a s s e r s a h e i n

t u v a s l i r e ? l^{'} h e u r e

Answered by mathmax by abdo last updated on 11/Jun/21

I = \int \sqrt{1 + e^{x}} dx we do the changement \sqrt{1 + e^{x}} = t \Rightarrow

1 + e^{x} = t^{2} \Rightarrow e^{x} = t^{2} - 1 \Rightarrow x = \log (t^{2} - 1) \Rightarrow

I = \int t \times \frac{2 t}{t^{2} - 1} dt = 2 \int \frac{t^{2}}{t^{2} - 1} dt = 2 \int \frac{t^{2} - 1 + 1}{t^{2} - 1} dt

= 2 t + \int \frac{2 dt}{t^{2} - 1} = 2 t + \int (\frac{1}{t - 1} - \frac{1}{t + 1}) dt = 2 t + \log ∣ \frac{t - 1}{t + 1} ∣ + C

I = 2 \sqrt{1 + e^{x}} + \log ∣ \frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x} + 1}} ∣ + C

Commented by mathmax by abdo last updated on 11/Jun/21

typo I = 2 \sqrt{1 + e^{x}} + \log ∣ \frac{\sqrt{1 + e^{x}} - 1}{\sqrt{1 + e^{x}} + 1} ∣ + C

Answered by puissant last updated on 23/Aug/21

u = \sqrt{e^{x} + 1} \to e^{x} = u^{2} - 1 \to x = l n (u^{2} - 1)

d x = \frac{2 u}{u^{2} - 1} d u

K = 2 \int \frac{u^{2}}{u^{2} - 1} d u

K = 2 \int d u + 2 \int \frac{1}{(u - 1) (u + 1)} d u

\Rightarrow K = 2 u + \int \frac{1}{u - 1} d u - \int \frac{1}{u + 1} d u

\Rightarrow K = 2 u + l n ∣ u - 1 ∣ - l n ∣ u + 1 ∣ + C

∴∵ K = 2 \sqrt{e^{x} + 1} + l n ∣ \frac{\sqrt{e^{x} + 1} - 1}{\sqrt{e^{x} + 1} + 1} ∣ + C . .

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