Solutions - 0050-02

(1) Trig substitution. Notice $x^2+4$, so we should make use of the identity ${\tan }^2\theta +1={\sec }^2\theta$.

Pick $x=2\tan \theta$ and thus $dx=2{\sec }^2\theta d\theta$.

Then:

$$\begin{array}{c}{x}^2+4\gg \gg 4{\tan }^2\theta +4\\ \\ \gg \gg 4({\tan }^2\theta +1)\gg \gg 4{\sec }^2\theta \end{array}$$

Plug in:

$$\int \frac{dx}{\sqrt{x^2+4}}\gg \gg \int \frac{2{\sec }^2\theta }{\sqrt{4{\sec }^2\theta }}d\theta \gg \gg \int \sec \theta d\theta$$

(We assume that $\sec \theta >0$ for the relevant values of $\theta$.)

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(2) Perform integration.

Either recall from memory, or multiply above and below by $\sec \theta +\tan \theta$, and obtain:

$$\int \sec \theta d\theta \gg \gg \ln |\tan \theta +\sec \theta |+C$$

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(3) To convert to $x$ we need $\sec \theta$ given that $\tan \theta =x/2$.

Draw triangle expressing $\tan \theta ={\displaystyle \frac{x}{2}}$:

Therefore $\sec \theta ={\displaystyle \frac{\sqrt{x^2+4}}{2}}$. We already know $\tan \theta ={\displaystyle \frac{x}{2}}$. Thus:

$$\int \sec \theta d\theta \gg \gg \ln |\frac{x}{2}+\frac{\sqrt{x^2+4}}{2}|+C$$

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(4) Simplify with log rules:

$$\begin{array}{c}\ln |\frac{x}{2}+\frac{\sqrt{x^2+4}}{2}|+C\gg \gg \ln \frac{1}{2}|x+\sqrt{x^2+4}|+C\\ \\ \gg \gg \ln \frac{1}{2}+\ln |x+\sqrt{x^2+4}|+C\gg \gg \ln |x+\sqrt{x^2+4}|+C\end{array}$$