What is the fast way to evaluate the following integral: $\int{\frac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$?

Question

I am trying to evaluate the following integral:

$$\int{\dfrac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$$

I tried the trigonometric substitution: $u = \tan(x)$. Generally, The whole integral needs two substitutions: $u = \tan(x)$ then $v = \sin(u)$. In order to get rid of trigonometric functions, one needs to know that: $$\sin(\arctan(x))=\dfrac{x}{\sqrt{x^2+1}}$$

My question is: What is the fast substitution that leads to the answer without passing by the above steps?

Answer 1

Rewrite as

$$\int dx \, x \frac1{x^4} \sqrt{1+\frac1{x^2}} = \frac12 \int du \frac1{u^2} \sqrt{1+\frac1{u}} = -\frac12 \int dv \sqrt{1+v}$$

In the above, $u=x^2$ and $v=1/u$. Thus, the antiderivative is

$$-\frac12 \cdot \frac23 (1+v)^{3/2} + C = -\frac13 \left (1+\frac1{x^2} \right )^{3/2}+C$$

Answer 2

We can simplify the expression $\frac{\sqrt{x^2+1}}{x^4}$, assuming that $x>0$, as follows: $$\frac{\sqrt{x^2+1}}{x^4} = \frac{\sqrt{1+\frac1{x^2}}}{x^3} = \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}}.$$ (We have multiplied both numerator and denominator by $\frac1x$.)

After this simplification $t=\frac1x$ seems as a reasonable substitution. It leads to $$\newcommand{\dd}{\; \mathrm{d}}\int \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}} \dd x = \begin{vmatrix} x=\frac1t & \dd x = -\frac{\dd t}{t^2} \\ t=\frac1x & \dd t = -\frac{\dd x}{x^2}\end{vmatrix} = -\int t\sqrt{1+t^2} \dd t.$$

The last integral is rather simple.

Hint: Try the substitution $u=1+t^2$.

Another reasonable substitution seems to be $t=\frac1{x^2}$, which leads to $$\int \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}} \dd x = \begin{vmatrix} t=\frac1{x^2} \\ \dd t=-\frac{\dd x}{x^3}\end{vmatrix} = -\int \sqrt{1+t} \dd t.$$

Answer 3

For clarifying the sign of the antiderivative, I decide to integrate it by cases:

A. When $x>0,$ $$ \begin{aligned} I &=\int \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} d x \\ &=-\frac{1}{2} \int \sqrt{1+\frac{1}{x^{2}}} d\left(1+\frac{1}{x^{2}}\right) \\ &=-\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{\frac{3}{2}}+C\\&=-\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3 x^{3}}+C \end{aligned} $$ B. When $x<0,$ let $y=-x$, then $$ I=\int \frac{\sqrt{y^{2}+1}}{y^{4}}(-d y)=-\int \frac{\sqrt{y^{2}+1}}{y^{4}} d y=\frac{\left(y^{2}+1\right)^{2}}{3 y^{3}}+C =-\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3 x^{3}}+C $$

Therefore we can conclude that $$I=-\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3 x^{3}}+C.$$