Integral of $$$\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}}$$$ with respect to $$$x$$$

Find $$$\int \frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}}\, dx$$$.

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,

$$$\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}} = \frac{1}{\left(a^{2} \sinh^{2}{\left( u \right)} + a^{2}\right)^{\frac{3}{2}}}$$$

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

$$$\frac{1}{\left(a^{2} \sinh^{2}{\left( u \right)} + a^{2}\right)^{\frac{3}{2}}}=\frac{\left|{a}\right|}{a^{4} \left(\sinh^{2}{\left( u \right)} + 1\right)^{\frac{3}{2}}}=\frac{\left|{a}\right|}{a^{4} \left(\cosh^{2}{\left( u \right)}\right)^{\frac{3}{2}}}$$$

$$$\frac{\left|{a}\right|}{a^{4} \left(\cosh^{2}{\left( u \right)}\right)^{\frac{3}{2}}} = \frac{\left|{a}\right|}{a^{4} \cosh^{3}{\left( u \right)}}$$$

Integral becomes

$${\color{red}{\int{\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}} d x}}} = {\color{red}{\int{\frac{1}{a^{2} \cosh^{2}{\left(u \right)}} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{a^{2}}$$$ and $$$f{\left(u \right)} = \frac{1}{\cosh^{2}{\left(u \right)}}$$$:

$${\color{red}{\int{\frac{1}{a^{2} \cosh^{2}{\left(u \right)}} d u}}} = {\color{red}{\frac{\int{\frac{1}{\cosh^{2}{\left(u \right)}} d u}}{a^{2}}}}$$

Rewrite the integrand in terms of the hyperbolic secant:

$$\frac{{\color{red}{\int{\frac{1}{\cosh^{2}{\left(u \right)}} d u}}}}{a^{2}} = \frac{{\color{red}{\int{\operatorname{sech}^{2}{\left(u \right)} d u}}}}{a^{2}}$$

The integral of $$$\operatorname{sech}^{2}{\left(u \right)}$$$ is $$$\int{\operatorname{sech}^{2}{\left(u \right)} d u} = \tanh{\left(u \right)}$$$:

$$\frac{{\color{red}{\int{\operatorname{sech}^{2}{\left(u \right)} d u}}}}{a^{2}} = \frac{{\color{red}{\tanh{\left(u \right)}}}}{a^{2}}$$

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

$$\frac{\tanh{\left({\color{red}{u}} \right)}}{a^{2}} = \frac{\tanh{\left({\color{red}{\operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}}} \right)}}{a^{2}}$$

Therefore,

$$\int{\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}} d x} = \frac{x}{a^{2} \sqrt{1 + \frac{x^{2}}{a^{2}}} \left|{a}\right|}$$

Simplify:

$$\int{\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}} d x} = \frac{x}{a^{2} \sqrt{a^{2} + x^{2}}}$$

Add the constant of integration:

$$\int{\frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}} d x} = \frac{x}{a^{2} \sqrt{a^{2} + x^{2}}}+C$$

$$$\int \frac{1}{\left(a^{2} + x^{2}\right)^{\frac{3}{2}}}\, dx = \frac{x}{a^{2} \sqrt{a^{2} + x^{2}}} + C$$$A