Integral of $$$\frac{\sqrt{x^{2} + 4}}{x^{4}}$$$

Find $$$\int \frac{\sqrt{x^{2} + 4}}{x^{4}}\, dx$$$.

Let $$$x=2 \sinh{\left(u \right)}$$$.

Then $$$dx=\left(2 \sinh{\left(u \right)}\right)^{\prime }du = 2 \cosh{\left(u \right)} du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{asinh}{\left(\frac{x}{2} \right)}$$$.

So,

$$$\frac{\sqrt{x^{2} + 4}}{x^{4}} = \frac{\sqrt{4 \sinh^{2}{\left( u \right)} + 4}}{16 \sinh^{4}{\left( u \right)}}$$$

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

$$$\frac{\sqrt{4 \sinh^{2}{\left( u \right)} + 4}}{16 \sinh^{4}{\left( u \right)}}=\frac{\sqrt{\sinh^{2}{\left( u \right)} + 1}}{8 \sinh^{4}{\left( u \right)}}=\frac{\sqrt{\cosh^{2}{\left( u \right)}}}{8 \sinh^{4}{\left( u \right)}}$$$

$$$\frac{\sqrt{\cosh^{2}{\left( u \right)}}}{8 \sinh^{4}{\left( u \right)}} = \frac{\cosh{\left( u \right)}}{8 \sinh^{4}{\left( u \right)}}$$$

So,

$${\color{red}{\int{\frac{\sqrt{x^{2} + 4}}{x^{4}} d x}}} = {\color{red}{\int{\frac{\cosh^{2}{\left(u \right)}}{4 \sinh^{4}{\left(u \right)}} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = \frac{\cosh^{2}{\left(u \right)}}{\sinh^{4}{\left(u \right)}}$$$:

$${\color{red}{\int{\frac{\cosh^{2}{\left(u \right)}}{4 \sinh^{4}{\left(u \right)}} d u}}} = {\color{red}{\left(\frac{\int{\frac{\cosh^{2}{\left(u \right)}}{\sinh^{4}{\left(u \right)}} d u}}{4}\right)}}$$

Multiply the numerator and denominator by $$$\frac{1}{\cosh^{4}{\left( u \right)}}$$$ and convert $$$\frac{\cosh^{4}{\left( u \right)}}{\sinh^{4}{\left( u \right)}}$$$ into $$$\frac{1}{\tanh^{4}{\left( u \right)}}$$$:

$$\frac{{\color{red}{\int{\frac{\cosh^{2}{\left(u \right)}}{\sinh^{4}{\left(u \right)}} d u}}}}{4} = \frac{{\color{red}{\int{\frac{1}{\cosh^{2}{\left(u \right)} \tanh^{4}{\left(u \right)}} d u}}}}{4}$$

Let $$$v=\tanh{\left(u \right)}$$$.

Then $$$dv=\left(\tanh{\left(u \right)}\right)^{\prime }du = \operatorname{sech}^{2}{\left(u \right)} du$$$ (steps can be seen »), and we have that $$$\operatorname{sech}^{2}{\left(u \right)} du = dv$$$.

The integral becomes

$$\frac{{\color{red}{\int{\frac{1}{\cosh^{2}{\left(u \right)} \tanh^{4}{\left(u \right)}} d u}}}}{4} = \frac{{\color{red}{\int{\frac{1}{v^{4}} d v}}}}{4}$$

Apply the power rule $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-4$$$:

$$\frac{{\color{red}{\int{\frac{1}{v^{4}} d v}}}}{4}=\frac{{\color{red}{\int{v^{-4} d v}}}}{4}=\frac{{\color{red}{\frac{v^{-4 + 1}}{-4 + 1}}}}{4}=\frac{{\color{red}{\left(- \frac{v^{-3}}{3}\right)}}}{4}=\frac{{\color{red}{\left(- \frac{1}{3 v^{3}}\right)}}}{4}$$

Recall that $$$v=\tanh{\left(u \right)}$$$:

$$- \frac{{\color{red}{v}}^{-3}}{12} = - \frac{{\color{red}{\tanh{\left(u \right)}}}^{-3}}{12}$$

Recall that $$$u=\operatorname{asinh}{\left(\frac{x}{2} \right)}$$$:

$$- \frac{\tanh^{-3}{\left({\color{red}{u}} \right)}}{12} = - \frac{\tanh^{-3}{\left({\color{red}{\operatorname{asinh}{\left(\frac{x}{2} \right)}}} \right)}}{12}$$

Therefore,

$$\int{\frac{\sqrt{x^{2} + 4}}{x^{4}} d x} = - \frac{2 \left(\frac{x^{2}}{4} + 1\right)^{\frac{3}{2}}}{3 x^{3}}$$

Simplify:

$$\int{\frac{\sqrt{x^{2} + 4}}{x^{4}} d x} = - \frac{\left(x^{2} + 4\right)^{\frac{3}{2}}}{12 x^{3}}$$

Add the constant of integration:

$$\int{\frac{\sqrt{x^{2} + 4}}{x^{4}} d x} = - \frac{\left(x^{2} + 4\right)^{\frac{3}{2}}}{12 x^{3}}+C$$

$$$\int \frac{\sqrt{x^{2} + 4}}{x^{4}}\, dx = - \frac{\left(x^{2} + 4\right)^{\frac{3}{2}}}{12 x^{3}} + C$$$A