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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?

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  • $\begingroup$ With Ron's method as a nice alternative, your approach is pretty much standard. I consider it fast enough as this integral isn't hard. Once you are well versed with certain identities and standard trig anti derivatives, you can do it halfway in the head... $\endgroup$ – imranfat Mar 13 '16 at 3:11
  • $\begingroup$ When you are a new teacher, you are learning yourself too. I have been teaching calculus for a decade and believe me, I am still learning new sleeky methods (for example on this website). In your situation my advice is to stick with the standard methods as these methods are to be understood by your students. Trig subs and trig back subs fall in that category. So your set up is the way for this integral. Once you are getting more "comfortable" in teaching the course, you may experiment with some alternative methods, depending on the "quality" of the batch of students you are having. $\endgroup$ – imranfat Mar 13 '16 at 3:20
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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

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

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  • $\begingroup$ Thanks, but this needs 3 substitutions: $u=x^2$, $v=1/u$ and $w=1+v$ $\endgroup$ – Navaro Mar 13 '16 at 2:21
  • $\begingroup$ @Navaro: I figured the last one was trivial enough to not need further mention. $\endgroup$ – Ron Gordon Mar 13 '16 at 2:21
  • $\begingroup$ Ok, Thank you so much :) $\endgroup$ – Navaro Mar 13 '16 at 2:22
  • $\begingroup$ You can save one step setting directly $\;v=\dfrac1{x^2}$, and $\mathrm d\mkern1mu v=-\dfrac2{x^3}\mathrm d\mkern 1mu x$. $\endgroup$ – Bernard Mar 13 '16 at 2:37
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    $\begingroup$ @Navaro: Here's the way I look at this. In this antiderivative computation, there is no range of $x$ given. So here I assumed that $x \gt 0$ for simplicity. If $x \lt 0$ then we may change the sign in the intermediate steps and get to the same result anyway. This is the way it goes in typical antiderivative calculations, so I do not worry about the sign like that, $\endgroup$ – Ron Gordon Mar 13 '16 at 3:08
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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 $$\int \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}} \dd x = \begin{vmatrix} t=\frac1{x^2} \\ \dd t=-\frac{\dd x}{x^3}\end{vmatrix} = -\int \sqrt{1+t} \dd t.$$

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