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Perform the following integration: $$ \int \dfrac{ 1}{x^2 \sqrt{x^2+1} } dx $$

Answer:

Let $I$ be the integral we are trying to evaluate. Let $\tan \theta = x$. We have: \begin{align*} dx &= \sec^2 \theta \,\, d\theta \\ I &= \int \dfrac{ 1}{\tan^2 \theta \sqrt{ \tan^2 \theta +1} } \,\, \sec^2 \theta \,\, d\theta \\ % I &= \int \dfrac{ 1}{\tan^2 \theta \sec \theta } \,\, \sec^2 \theta \,\, d\theta \\ % I &= \int \dfrac{ 1}{\tan^2 \theta } \,\, \sec \theta \,\, d\theta \\ % I &= \int \left(\dfrac{1}{ \cos \theta } \right) \left( \dfrac{ \cos^2 \theta }{ \sin^2 \theta } \right) \,\, d\theta \\ % I &= \int \cot \theta \csc \theta \,\, d\theta \\ I &= - \csc \theta + C \\ \end{align*} Now we need to substitute back. \begin{align*} I &= -\dfrac{1}{\sin \theta} + C \\ \end{align*} How do I finish the problem?
Using an online integral calculator I find the answer is: $$ I = - \dfrac{ \sqrt{ x^2 + 1 } }{x} $$

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    $\begingroup$ Continuing your work, you just need to express $\csc\theta$ in terms of $x$. Recall that $\csc^2\theta=\cot^2\theta+1$ and hence $\csc\theta=\sqrt{\cot^2\theta+1}=\sqrt{(1/x)^2+1}$, now simplify. Do note that this all assumes $x\gt 0$ and we simplify the square roots $\sqrt{(\cdots)^2}$ as $\cdots$ instead of $|\cdots|$ $\endgroup$ Commented Sep 8, 2023 at 21:09
  • $\begingroup$ Almost any integral is as described in the title. $\endgroup$ Commented Sep 8, 2023 at 23:07

8 Answers 8

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When $x>0$ we have

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

$$ = \int \frac{d\left(\frac{1}{x^2}\right)}{-2\sqrt{1+\frac{1}{x^2}}} = -\sqrt{1+\frac{1}{x^2}}+C$$

Pulling the $x^{-2}$ factor out allows us to say that the antiderivative is

$$-\frac{\sqrt{x^2+1}}{x} + C$$

for all $x\neq 0$.

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  • $\begingroup$ You didn't work out for $x<0$? $\endgroup$ Commented Aug 10 at 14:45
  • $\begingroup$ @Integreek don't need to. You can see that when the final answer before pulling out the factor would have had an additional minus sign, we would represent that as a hidden factor of $\operatorname{sgn}(x)$. Pulling out the factor from the square root, the denominator becomes $\operatorname{sgn}(x)\cdot\sqrt{x^2} = \operatorname{sgn}(x)\cdot|x| = x$ $\endgroup$ Commented Aug 11 at 3:21
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Here's how to get to the finish line using the substitution.

From $x = \tan \theta$, you can construct a right triangle whose hypotenuse has length $\sqrt{1 + x^2}.$ (This is just from using Pythagoras and your basic trigonometric ratios). Now based on that picture you've drawn. $\sin \theta = \ldots$ (finish this).

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$$ \int \frac{1}{x^2\sqrt{x^2+1}}dx = [u =-\frac{1}{x}] = \int \frac{u}{\sqrt{1+u^{2}}}du = \sqrt{1+u^{2}} +C= -\frac{\sqrt{x^2+1}}{x}+C. $$

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  • $\begingroup$ I thought about this way too. $\endgroup$ Commented Sep 9, 2023 at 5:42
  • $\begingroup$ But there is a sign problem... $\endgroup$ Commented Sep 9, 2023 at 5:54
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Another possible approach based on hyperbolic functions: \begin{align*} \int\frac{\mathrm{d}x}{x^{2}\sqrt{1 + x^{2}}} & = \int\frac{\mathrm{d}(\sinh(u))}{\sinh^{2}(u)\sqrt{1 + \sinh^{2}(x)}}\\\\ & = \int\frac{\cosh(u)}{\sinh^{2}(u)\cosh(u)}\mathrm{d}u\\\\ & = \int\frac{\mathrm{d}u}{\sinh^{2}(u)}\\\\ & = \int\operatorname{csch}^{2}(u)\mathrm{d}u\\\\ & = -\coth(u) + C\\\\ & = -\frac{\cosh(u)}{\sinh(u)} + C\\\\ & = -\frac{\sqrt{1 + \sinh^{2}(u)}}{\sinh(u)} + C\\\\ & = -\frac{\sqrt{1 + x^{2}}}{x} + C \end{align*}

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Substitute $y= \sqrt{x^2+1} $. Then, $y’=\frac xy$ \begin{align} \int &\frac{dx}{x^2\sqrt{x^2+1}} = \int \frac{dx}{x^2 y} =\int\left(y-\frac{x^2}y\right)\frac{dx}{x^2}\\ &=\int\left(\frac{y}{x^2}-\frac{y’}x\right)dx =-\int d\left( \frac yx \right)=-\frac yx+C \end{align}

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You got: $I=-\dfrac{1}{\sin \theta}+C$, where $\theta = \arctan x$.

Hence, it is: $$I=-\frac{1}{\sin(\arctan x)}+C= -\frac{\dfrac{1}{1}}{\dfrac{\tan(\arctan x)}{\sqrt{1+\tan^2(\arctan x)}}}+C=\boxed{-\frac{\sqrt{1+x^2}}{x}+C}.$$

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My Solution is similar to Hunaphu's solution. I considered sign (negative/positive) situations.

İf we sub $x=\frac1t$ in $\int\frac{dx}{x^2\sqrt{x^2+1}}$ we get, $\begin{align}\int-\frac{|t|}{\sqrt{t^2+1}}dt&=\int-sgn(t)\frac{t}{\sqrt{t^2+1}}dt\\ &=-sgn(t)\sqrt{t^2+1}+c\\ &=-sgn(t)\frac{\sqrt{x^2+1}}{|x|}+c\\ &=-\frac{sgn(t)}{sgn(x)}\frac{\sqrt{x^2+1}}{x}+c\\ &=-\frac{\sqrt{x^2+1}}{x}+c \end{align}$

Since $sgn(t)=sgn(x)$.

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You can directly read off $$ \csc\theta = \frac{\text{hypotenuse}}{\text{opposite side}}$$ as a trigonometric ratio in a right triangle with proper sign:

enter image description here

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