How do you evaluate the integral int sqrt(e^x+1)?

May 23, 2018

Use the substitution $\sqrt{{e}^{x} + 1} = u$.

Explanation:

Let

$I = \int \sqrt{{e}^{x} + 1} \mathrm{dx}$

Apply the substitution $\sqrt{{e}^{x} + 1} = u$:

$I = 2 \int \frac{{u}^{2}}{{u}^{2} - 1} \mathrm{du}$

Rearrange:

$I = 2 \int \left(1 + \frac{1}{{u}^{2} - 1}\right) \mathrm{du}$

Factorize the denominator:

$I = 2 \int \left(1 + \frac{1}{\left(u - 1\right) \left(u + 1\right)}\right) \mathrm{du}$

Apply partial fraction decomposition:

$I = \int \left(2 + \frac{1}{u - 1} - \frac{1}{u + 1}\right) \mathrm{du}$

Integrate term by term:

$I = 2 u + \ln | u - 1 | - \ln | u + 1 | + C$

Reverse the substitution:

$I = 2 \sqrt{{e}^{x} + 1} + \ln | \frac{\sqrt{{e}^{x} + 1} - 1}{\sqrt{{e}^{x} + 1} + 1} | + C$