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Evaluate integral with respect to x (that's what I call it)

Integral is given as

Square root of x-square plus 1
whole divided by
x raise to 4

$$\int{\sqrt{x^2+1}\over x^4}\,dx$$

sorry i don't no formatting

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closed as off-topic by user127.0.0.1, Davide Giraudo, abiessu, TooTone, Your Ad Here Mar 3 '14 at 17:18

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If this question can be reworded to fit the rules in the help center, please edit the question.

Please, tell us what you tried and where you have problems. –  Claude Leibovici Mar 3 '14 at 12:06
Integrals involving $\sqrt{x^2+1}$ can often be done using $x=\tan u$, $dx=\sec^2u$. –  Gerry Myerson Mar 3 '14 at 12:10
Here are some examples of integrals where the substitution suggested by Gerry Myerson was used. (Although they seem to be a little simpler than this one.) math.stackexchange.com/questions/591073/… and math.stackexchange.com/questions/179129/… –  Martin Sleziak Mar 3 '14 at 12:52
For some basic information about writing math at this site see e.g. here, here, here and here. –  Martin Sleziak Mar 3 '14 at 14:06

2 Answers 2

As suggested in the comments, the substitution $x=\tan u$, $dx = \frac{du}{\cos^2 u}$ leads to $$ \int \frac{\cos u}{\sin^4 u}du = -\frac{1}{3} \frac{1}{\sin^3 u}. $$ Now you can express $\sin u$ as a function of $\tan u$, and get back to the variable $x$.

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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 $u=1+t^2$.

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