Evaluate the indefinite integral of e^{x^1/2}

by RoRi

November 11, 2015

Compute the following indefinite integral

$\int e^{\sqrt{x}} \, dx.$

To evaluate this integral we want to make a substitution. First multiply the numerator and denominator by $2 \sqrt{x}$ to obtain,

$\int e^{\sqrt{x}} \, dx = \int \frac{2\sqrt{x} e^{\sqrt{x}}}{2 \sqrt{x}} \, dx.$

Now define the function $u$ by

$u = e^{\sqrt{x}}, \qquad du &= \frac{e^{\sqrt{x}}}{2\sqrt{x}}.$

This implies

$\log u = \sqrt{x}.$

Therefore, using the method of substitution, we have

$\begin{align*} \int e^{\sqrt{x}} \, dx &= \int 2 \log u \, du \\ &= 2 (u \log u - u) + C \\ &= 2 \left( e^{\sqrt{x}} \log (e^{\sqrt{x}}) - e^{\sqrt{x}} \right) + C\\ &= 2 (\sqrt{x} - 1)e^{\sqrt{x}} + C. \end{align*}$

Stumbling Robot

Evaluate the indefinite integral of e^{x^1/2}

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