# How do you find the antiderivative of (1 + e^(2x)) ^(1/2)?

May 8, 2017

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

#### Explanation:

Let us subst. ${e}^{x} = \tan y \Rightarrow {e}^{x} \mathrm{dx} = {\sec}^{2} y \mathrm{dy} , \mathmr{and} , \mathrm{dx} = {\sec}^{2} \frac{y}{e} ^ x \cdot \mathrm{dy} = {\sec}^{2} \frac{y}{\tan} y \cdot \mathrm{dy} = \frac{1}{\cos y \sin y} \mathrm{dy}$

$\therefore I = \int \sqrt{1 + {e}^{2 x}} \mathrm{dx}$

$= \int \left\{\frac{\sqrt{1 + {\tan}^{2} y}}{\cos y \sin y}\right\} \mathrm{dy} ,$

$= \int \frac{1}{{\cos}^{2} y \sin y} \mathrm{dy} = \int \left\{\frac{\sin y}{{\cos}^{2} y {\sin}^{2} y}\right\} \mathrm{dy} .$

Hence, $\cos y = t \Rightarrow - \sin y \mathrm{dy} = \mathrm{dt} , \mathmr{and} , \therefore ,$

$I = - \int \frac{1}{{t}^{2} \left(1 - {t}^{2}\right)} \mathrm{dt} ,$

$= \int \frac{1}{{t}^{2} \left({t}^{2} - 1\right)} \mathrm{dt} = \int \left[\frac{{t}^{2} - \left({t}^{2} - 1\right)}{{t}^{2} \left({t}^{2} - 1\right)}\right] \mathrm{dt} ,$

$= \int \left[{t}^{2} / \left\{{t}^{2} \left({t}^{2} - 1\right)\right\} - \frac{{t}^{2} - 1}{{t}^{2} \left({t}^{2} - 1\right)}\right] \mathrm{dt} ,$

$= \int \left(\frac{1}{{t}^{2} - 1} - \frac{1}{t} ^ 2\right) \mathrm{dt} ,$

$= \frac{1}{2} \ln | \frac{t - 1}{t + 1} | + \frac{1}{t} ,$

$= \frac{1}{2} \ln | \frac{\cos y - 1}{\cos y + 1} | + \frac{1}{\cos} y ,$

$= \frac{1}{2} \ln | \frac{1 - \sec y}{1 + \sec y} | + \sec y .$

Since, $\tan y = {e}^{x} \Rightarrow \sec y = \sqrt{1 + {\tan}^{2} y} = \sqrt{1 + {e}^{2 x}} ,$ we get,

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

Enjoy Maths.!