Evaluate the integral of (x2 + 5)1/2

Compute the following integral.

$\int \sqrt{x^2+5} \, dx.$

First, we want to make the substitution

$\begin{align*} x &= \sqrt{5} \tan s & \implies \qquad dx &= \sqrt{5} \sec^2 s \, ds \\ x^2 &= 5 \tan^2 s. \end{align*}$

Therefore, we have

$\begin{align*} \int \sqrt{x^2+5} \, dx &= \int \sqrt{5 \tan^2 s + 5} (\sqrt{5} \sec^2 s) \, ds \\[9pt] &= 5 \int \sqrt{\tan^2 s + 1} \sec^2 s \, ds \\[9pt] &= 5 \int \sec^3 s \, ds. \end{align*}$

Now, we use integration by parts, letting

$\begin{align*} u &= \sec s & \implies && du &= \sec s \tan s \, ds \\ dv &= \sec^2 s \, ds & \implies && v &= \tan s. \end{align*}$

Therefore, we have

$\begin{align*} \int \sec^3 s \, ds &= \sec s \tan s - \int \tan^2 s \sec s \, ds \\[9pt] &= \sec s \tan s - \int \sec s (\sec^2 s -1) \, ds \\[9pt] &= \sec s \tan s - \int \sec^3 s \, ds + \int \sec s \, ds. \end{align*}$

Moving the $\int \sec^3 s \, ds$ back to the left side we have

$\begin{align*} && 2 \int \sec^3 s \, ds &= \sec s \tan s + \int \sec s \, ds \\[9pt] \implies && \int sec^3 s \, ds &= \frac{1}{2} \sec s \tan s + \frac{1}{2} \int \sec s \, ds \\[9pt] &&&= \frac{1}{2} \sec s \tan s + \frac{1}{2} \log |\sec s + \tan s| + C. \end{align*}$

Plugging this back in above we have,

$\begin{align*} \int \sqrt{x^2+5} \, dx &= 5 \int \sec^3 s \, ds \\[9pt] &= \frac{5}{2} \sec s \tan s + \frac{5}{2} \log | \sec s + \tan s| + C \\[9pt] &= \frac{5}{2} \sec \left( \arctan \frac{x}{\sqrt{5}} \right) \frac{x}{\sqrt{5}} + \frac{5}{2} \log \left| \sec \left( \arctan \frac{x}{\sqrt{5}} \right) + \frac{x}{\sqrt{5}} \right| + C. \end{align*}$

Then we note that

$\sec \left( \arctan \frac{x}{\sqrt{5}} \right) = \frac{\sqrt{x^2+5}}{\sqrt{5}}.$

(If you don’t remember your trigonometry, which I usually don’t, you can figure things like this out by drawing the triangle. Since $\theta = \arctan \frac{x}{\sqrt{5}}$ means you have a right triangle with legs of length $x$ and $\sqrt{5}$ . That means the hypotenuse has length $\sqrt{x^2+5}$ . So, $\cos \theta = \frac{x}{\sqrt{x^2+5}}$ which then gives the above result for $\sec \theta$ .)
Plugging this back in we then have

$\begin{align*} \int \sqrt{x^2+5} \, dx &= \frac{5}{2} \sec \left( \arctan \frac{x}{\sqrt{5}} \right) \frac{x}{\sqrt{5}} + \frac{5}{2} \log \left| \sec \left( \arctan \frac{x}{\sqrt{5}} \right) + \frac{x}{\sqrt{5}} \right| + C \\[9pt] &= \frac{x}{2} \sqrt{x^2+5} + \frac{5}{2} \log \left| \frac{x + \sqrt{x^2+5}}{\sqrt{5}} \right| + C \\[9pt] &= \frac{x}{2} \sqrt{x^2+5} + \frac{5}{2} \log \left| x + \sqrt{x^2+5} \right| - \frac{5}{2} \log \sqrt{5} + C \\[9pt] &= \frac{x}{2} \sqrt{x^2+5} + \frac{5}{2} \log \left| x + \sqrt{x^2+5} \right| + C. \end{align*}$

(Where we absorbed $\frac{5}{2} \log \sqrt{5}$ into the constant in the final line, since it is just a constant.)

Stumbling Robot

Evaluate the integral of (x² + 5)^1/2

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