# Evaluate these integrals. intsqrt{e^x+1}dx

Evaluate these integrals.
$\int \sqrt{{e}^{x}+1}dx$
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Delorenzoz

Evaluate:
$\int \sqrt{{e}^{x}+1}dx$ (1)
Simplification:
Substitute:
${e}^{x}+1=t$
${e}^{x}dx=dt$
$dx=\frac{dt}{{e}^{x}}$
$dx=\frac{dt}{t-1}$
in(1),
$\int \sqrt{{e}^{x}+1}dx=\int \frac{\sqrt{t}}{t-1}dt$ (2)
Now again substitute
$\sqrt{t}=u$
$\left[\frac{1}{2\sqrt{t}}dt=du⇒dt=2\sqrt{t}du⇒dt=2{u}^{2}du\right]$
in (2),
$\int \sqrt{{e}^{x}+1}dx=\int \frac{\sqrt{t}}{t-1}dt$
$=2\int \frac{u}{{u}^{2}-1}\left(2u\right)du$
$=2\int \frac{{u}^{2}}{{u}^{2}-1}du$
$=2\int \frac{{u}^{2}-1+1}{{u}^{2}-1}du$
$=2\int \left(\frac{{u}^{2}-1}{{u}^{2}-1}+\frac{1}{{u}^{2}-1}\right)du$
$=2\int \left(1+\frac{1}{{u}^{2}-1}\right)du$
$=2\int \left(1+\frac{1}{\left(u+1\right)\left(u-1\right)}\right)du$ (3)
Now evaluate $\frac{1}{\left(u+1\right)\left(u-1\right)}$ using partial fraction method,
$\frac{1}{\left(u+1\right)\left(u-1\right)}=\frac{A}{u+1}+\frac{B}{u-1}$
$1=A\left(u-1\right)+B\left(u+1\right)$

$\frac{1}{\left(u+1\right)\left(u-1\right)}=-\frac{1}{2\left(u+1\right)}+\frac{1}{2\left(u-1\right)}$ (4)
Now using (4) in (3),
$\int \sqrt{{e}^{x}+1}dx=2\int \left(1+\left(-\frac{1}{2\left(u+1\right)}+\frac{1}{2\left(u-1\right)}\right)\right)du$
$=2\int \left(1-\frac{1}{2\left(u+1\right)}+\frac{1}{2\left(u-1\right)}\right)du$
$=2u-\mathrm{ln}|u+1|+\mathrm{ln}|u-1|+C$
$=2\sqrt{t}-\mathrm{ln}|\sqrt{t}+1|+\mathrm{ln}|\sqrt{t}-1|+C$
$=2\sqrt{{e}^{x}+1}-\mathrm{ln}|\sqrt{{e}^{x}+1}+1|+\mathrm{ln}|\sqrt{{e}^{x}+1}-1|+C$
Hence
$\int \sqrt{{e}^{x}+1}dx=2\sqrt{{e}^{x}+1}-\mathrm{ln}|\sqrt{{e}^{x}+1}+1|+\mathrm{ln}|\sqrt{{e}^{x}+1}-1|+C$