# How do you evaluate the integral int e^sqrtx?

Feb 5, 2017

$\int {e}^{\sqrt{x}} \mathrm{dx} = 2 {e}^{\sqrt{x}} \left(\sqrt{x} - 1\right) + C$

#### Explanation:

Substitute:

$t = {e}^{\sqrt{x}}$

$x = {\left(\ln t\right)}^{2}$

$\mathrm{dx} = 2 \ln \frac{t}{t} \mathrm{dt}$

we have:

$\int {e}^{\sqrt{x}} \mathrm{dx} = 2 \int t \ln \frac{t}{t} \mathrm{dt} = 2 \int \ln t \mathrm{dt}$

we can now integrate by parts:

$\int \ln t \mathrm{dt} = t \ln t - \int t d \left(\ln t\right) = t \ln t - \int t \frac{\mathrm{dt}}{t} = t \ln t - \int \mathrm{dt} = t \ln t - t + C = t \left(\ln t - 1\right) + C$

undoing the substitution:

$\int {e}^{\sqrt{x}} \mathrm{dx} = 2 {e}^{\sqrt{x}} \left(\sqrt{x} - 1\right) + C$

Feb 5, 2017

Please see below for an alternative solution.

#### Explanation:

In order to integrate ${e}^{\sqrt{x}} \mathrm{dx}$ by substitution, we would need the derivative of $\sqrt{x}$. We will introduce the derivative and see if that helps. (If it doesn't help, we'll try something else.)

$\int {e}^{\sqrt{x}} \mathrm{dx} = \int {\underbrace{2 \sqrt{x}}}_{u} {\underbrace{{e}^{\sqrt{x}} / \left(2 \sqrt{x}\right) \mathrm{dx}}}_{\mathrm{dv}}$

With $u = 2 \sqrt{x}$ and $\mathrm{dv} = {e}^{\sqrt{x}} / \left(2 \sqrt{x}\right) \mathrm{dx}$, we can find

$\mathrm{du} = \frac{1}{\sqrt{x}}$ and (integrating by substitution) $v = {e}^{\sqrt{x}}$

$u v - \int v \mathrm{du} = 2 \sqrt{x} {e}^{\sqrt{x}} - \int {e}^{\sqrt{x}} / \sqrt{x} \mathrm{dx}$

$= 2 \sqrt{x} {e}^{\sqrt{x}} - 2 \int {e}^{\sqrt{x}} / \left(2 \sqrt{x}\right) \mathrm{dx}$

$= 2 \sqrt{x} {e}^{\sqrt{x}} - 2 {e}^{\sqrt{x}} + C$

$= 2 {e}^{\sqrt{x}} \left(\sqrt{x} - 1\right) + C$