Evaluate indefinite integral: $\int {\frac{\sqrt{1+z^2}}{z^3}dz}$

Question

Evaluate indefinite integral: $\int {\frac{\sqrt{1+z^2}}{z^3}dz}$

My try,

Substituting $z=\tan x$ then the integration reduces in $\int{\csc{^3} xdx}$. Where, $\csc x$ is Cosecant function.

After rewriting the integral as $\int {\csc x\csc^2xdx}$ and by parts and calculation I find the integral value $-\frac{1}2\csc x \cot x +\frac{1}{2}\ln|\csc x-\cot x|+c$

Now converting the the result in terms of $z$ is bit problematic.

I am finding a alternative way to Integrate it without substituting (if possible then without by parts). Does anyone have any idea how to do this? Your help is highly solicited.

Answer 1

Putting, $z=\frac{1}t$

$$\begin{aligned} \int \frac{\sqrt{1+z^2}}{z^3}dz &= \int {\frac{\sqrt {1+\frac{1}{t^2}}}{\frac{1}{t^3}}(-\frac{1}{t^2}dt)}= \left[ \begin{array}{l} z = \frac{1}{t} \Leftrightarrow dz = -\frac{1}{t^2}dt \\ \end{array} \right]\\ &= -\int {\sqrt{1+t^2}dt} = -\frac{1}{2}[t\sqrt {1+t^2}+\ln|t+\sqrt{1+t^2}|]+c \end{aligned}$$

After rewriting in terms of $z$ we get, $-\frac{1}{2}[\frac{\sqrt {1+z^2}}{z^2}+\ln|\frac{1}{z}+\frac{\sqrt {1+z^2}}{z}|]+c$

Answer 2

This is an example of a differentil binomial integration:

$$ \int x^m\left(a+bx^n\right)^pdx. $$

In your example,

$$ \int \frac{\sqrt{1+z^2}}{z^3}dz = \int z^{-3}\left(1+1\cdot z^2\right)^\frac{1}{2}dz, $$

i.e., $m = -3, n = 2, p = \frac{1}{2}, a = 1, b = 1$.

The integral can be evaluated in terms of the elementary functions if one of the following is true:

1. $m$ and $n$ are fractions, and $p$ is an integer. (not our case).
2. $p$ is a fraction and $\frac{m+1}{n}$ is an integer. (our case since $\frac{-3+1}{2} = -1$).
3. $p$ is a fraction and $p + \frac{m+1}{n}$ is an integer. (not our case).

So case 2 is true, and we have to do a substitution:

$$ a+bx^n = t^s, \text{ where }s\text{ is a denominator of }p, \text{ i.e., } s=2. $$ \begin{aligned} \int \frac{\sqrt{1+z^2}}{z^3}dz &= \int z^{-3}\left(1+1\cdot z^2\right)^\frac{1}{2}dz = \left[ \begin{array}{l} 1+z^2 = t^2 \Leftrightarrow t = \sqrt{1+z^2} \\ 2zdz = 2tdt \Leftrightarrow zdz = tdt \end{array} \right] = \\ &= \int \frac{\sqrt{1+z^2}}{z^4}zdz = \int \frac{t}{(t^2-1)^2}tdt = \int \frac{t^2}{(t^2-1)^2}dt. \end{aligned}

The latter integral can be evaluated by using partial fraction decomposition:

$$ \frac{t^2}{(t^2-1)^2} = -\frac{1}{4}\frac{1}{(t+1)}+\frac{1}{4}\frac{1}{(t-1)}+\frac{1}{4}\frac{1}{(t-1)^2}+\frac{1}{4}\frac{1}{(t+1)^2}. $$