Integral of $$$\frac{1}{\sqrt{a^{2} + x^{2}}}$$$ with respect to $$$x$$$

Find $$$\int \frac{1}{\sqrt{a^{2} + x^{2}}}\, dx$$$.

Let $$$x=\sinh{\left(u \right)} \left|{a}\right|$$$.

Then $$$dx=\left(\sinh{\left(u \right)} \left|{a}\right|\right)^{\prime }du = \cosh{\left(u \right)} \left|{a}\right| du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}$$$.

So,

$$$\frac{1}{\sqrt{a^{2} + x^{2}}} = \frac{1}{\sqrt{a^{2} \sinh^{2}{\left( u \right)} + a^{2}}}$$$

Use the identity $$$\sinh^{2}{\left( u \right)} + 1 = \cosh^{2}{\left( u \right)}$$$:

$$$\frac{1}{\sqrt{a^{2} \sinh^{2}{\left( u \right)} + a^{2}}}=\frac{1}{\sqrt{\sinh^{2}{\left( u \right)} + 1} \left|{a}\right|}=\frac{1}{\sqrt{\cosh^{2}{\left( u \right)}} \left|{a}\right|}$$$

$$$\frac{1}{\sqrt{\cosh^{2}{\left( u \right)}} \left|{a}\right|} = \frac{1}{\cosh{\left( u \right)} \left|{a}\right|}$$$

Therefore,

$${\color{red}{\int{\frac{1}{\sqrt{a^{2} + x^{2}}} d x}}} = {\color{red}{\int{1 d u}}}$$

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$${\color{red}{\int{1 d u}}} = {\color{red}{u}}$$

Recall that $$$u=\operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}$$$:

$${\color{red}{u}} = {\color{red}{\operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}}}$$

Therefore,

$$\int{\frac{1}{\sqrt{a^{2} + x^{2}}} d x} = \operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}$$

Add the constant of integration:

$$\int{\frac{1}{\sqrt{a^{2} + x^{2}}} d x} = \operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}+C$$

$$$\int \frac{1}{\sqrt{a^{2} + x^{2}}}\, dx = \operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)} + C$$$A