# Evaluating $\int \frac{\sqrt{1+x^2}}{x^3}dx$

Evaluating $\int \frac{\sqrt{1+x^2}}{x^3} dx$. Any suggestions? I thought the replacement $t=\sinh x$ but in this way should solve the integral of $\frac{\cosh^2 t}{\sinh^3 t}$ (more difficult for me).

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$\def\csch{\mathop{\mathrm{csch}}}\cosh^2 t / \sinh^3 t = \coth^2 t \csch t = \csch t + \csch^3 t$ –  Hurkyl Jun 25 '12 at 0:13

If you look at the form of the integrand, you can see it is

$$\sqrt{1+x^2}$$

Which is similar to our pythagorean identity

$$\sin^2(x) + \cos^2(x) = 1$$

If you fiddle around with the equation, you can obtain

$$\tan^2(x) + 1 = \sec^2(x)$$

$$\tan^2(x) + 1$$

looks just like

$$\sqrt{1+x^2}$$

So, we can make the substitution

$$x = \tan(u)$$

$$\sqrt{1 + x^2} = \sqrt{1 + \tan^2(x)} = \sqrt{\sec^2(x)} = \sec(x)$$

Substituting in for your original problem, we get

$$\int \frac{\sec(x)}{\tan^3(x)}\, dx$$

From here, I would simplify by writing in terms of sine and cosine and then solve. Remember that once you arrive at your answer, it will be in terms of u, in this case. You must draw a triangle with angle u, and since you have x = tan(u), you must derive from your triangle and substitute everything back in terms of x, your original variable.

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Put $t= \tan\theta$ , then what you get is $$\int \frac{\tan\theta}{\tan^{3}\theta} \cdot \frac{1}{\cos^{2}\theta} \ d\theta = \int \csc^{2}\theta \ d \theta$$
Just a nitpick, maybe you mean $x=\tan \theta$. Likewise $\sqrt{1+\tan^2 \theta}=\sec^2\theta\ne\tan\theta$ –  E.O. Jun 24 '12 at 13:19
Rewrite the integrand as follows: $$I=\int \frac{\sqrt{1+x^2}}{x^3}dx=\int\left(\sqrt{1+\frac{1}{x^2}}\right)\frac{dx}{x^2}=-\int\left(\sqrt{1+\frac{1}{x^2}}\right)d\left(\frac{1}{x}\right)$$ Now let $t=\frac{1}{x}$ $$I=-\int\sqrt{1+t^2}dt$$ The latter is a table integral (or you can let $x=\sinh t$ now)