Integral $\int \sqrt{x^{2}+a^{2}}$

Question

I want to determinate the following integral

$$\int \sqrt{x^{2}+a^{2}}dx \space | \space a> 0$$

I used integration by partition and u-substitution but I came to no result.

Answer 1

Let $x = a \tan \theta$. Then $dx = a \sec^2 \theta d \theta$, and you have:

$$\begin{align} \int \sqrt{(a \tan \theta)^2 + a^2}(a \sec^2 \theta) d \theta &= \int \sqrt{a^2(\tan^2 \theta + 1)}(a \sec^2 \theta) d \theta \\ &= \int (a \sec \theta)(a \sec^2 \theta) d \theta \\ &= a^2 \int \sec^3 \theta d \theta \end{align}$$

Now integrate by parts: $$\begin{align} u &= \sec \theta &dv &= \sec^2 \theta d\theta \\ du &= \sec \theta \tan \theta d \theta &v &= \tan \theta \end{align}$$

$$\begin{align} \int \sec^3 \theta d \theta &= \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta d \theta \\ &= \sec \theta \tan \theta - \int \sec \theta (\sec^2 \theta - 1) d \theta \\ &= \sec \theta \tan \theta - \int \sec^3 \theta d \theta + \int \sec \theta d \theta \\ \Rightarrow 2 \int \sec^3 \theta d \theta &= \sec \theta \tan \theta + \int \sec \theta d \theta \\ \Rightarrow \int \sec^3 \theta d \theta &= \frac{1}{2}\left(\sec \theta \tan \theta + \ln \left|\sec \theta + \tan \theta\right|\right)+C \end{align}$$

So then you have:

$$\begin{align} \int \sqrt{x^2 + a^2} dx &= a^2 \int \sec^3 \theta d \theta \\ &= \frac{a^2}{2}\left(\sec \theta \tan \theta + \ln|\sec \theta + \tan \theta|\right)+C \end{align}$$

Now since $\tan \theta = \frac{x}{a}$, and $\sec \theta = \frac{\sqrt{x^2 + a^2}}{a}$, you have:

$$\begin{align} \int \sqrt{x^2 + a^2} dx &= \frac{a^2}{2}\left(\frac{\sqrt{x^2 + a^2}}{a} \frac{x}{a} + \ln \left|\frac{\sqrt{x^2 + a^2}}{a} + \frac{x}{a}\right|\right)+C \\ &= \frac{1}{2} \left(x\sqrt{x^2 + a^2} + a^2 \ln \left|\sqrt{x^2 + a^2} + x\right|\right)+C' \end{align}$$

Answer 2

HINT: Try trigonometric substitution by letting $x = a\tan \theta \implies dx = a\sec^2 \theta$.

You can then use the fact that $\tan^2 \theta + 1 = \sec^2 \theta$.

Answer 3

Using integration by parts, you simply have$$\int \sqrt{x^{2}+a^{2}}dx =\int \sqrt{x^{2}+a^{2}}\cdot 1dx \\I= \sqrt{x^{2}+a^{2}}\cdot x-\int x\cdot \frac{x}{\sqrt{x^{2}+a^{2}}} dx \\ I= \sqrt{x^{2}+a^{2}}\cdot x-\int \frac{x^2+a^2-a^2}{\sqrt{x^{2}+a^{2}}} dx \\ I= \sqrt{x^{2}+a^{2}}\cdot x-I+a^2\int \frac{1}{\sqrt{x^{2}+a^{2}}} dx \\ 2I= \sqrt{x^{2}+a^{2}}\cdot x+a^2\int \frac{1}{x+\sqrt{x^{2}+a^{2}}}\cdot \frac{x+\sqrt{x^{2}+a^{2}}}{\sqrt{x^{2}+a^{2}}} dx \\ 2I= \sqrt{x^{2}+a^{2}}\cdot x+a^2\int \frac{1}{x+\sqrt{x^{2}+a^{2}}}\cdot( \frac{x}{\sqrt{x^{2}+a^{2}}}+1) dx \\ 2I= \sqrt{x^{2}+a^{2}}\cdot x+a^2\ln (x+\sqrt{x^{2}+a^{2}})+c$$