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I need to solve this integral:

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

First I thought it was easy, so I tried integration by parts with $g(x)=x$ and $g'(x)=1$:

$$\int{ \frac{x^2}{(1+x^2)^{\frac{3}{2}} }}\,\mathrm dx $$

But I've made it even more complicated than before, and if I want to solve it again by parts I'll have $g(x)= \frac{x^3}{3}$ , and I will never end integrating.

How should I solve it?

Edit

Trying this way: $x= \tan(t)$, then I get:

$$ \int{ \frac{1+\tan^2(t)}{ \sqrt{1+ \tan^2(t)} } \mathrm dt}= \int{ \sqrt{ 1 + \tan^2(t) } \, \mathrm dt }$$

But it doesn't remind me anything, I still can't solve it.

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    $\begingroup$ Just put $x=\sinh t=\frac{e^t-e^{-t}}{2}$ $\endgroup$ Commented Jul 4, 2013 at 19:29
  • $\begingroup$ Please, try to make the title of your question more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. $\endgroup$ Commented Jul 4, 2013 at 19:56
  • $\begingroup$ I first thought that $tg(t)$ means $t\times g(t)$ where $g(t)$ is some function in $t$ and was confused for $3$ minutes...then it struck me that you are actually referring to $\tan t$. $\endgroup$ Commented Feb 18 at 17:00

5 Answers 5

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When you see $1+x^2$ you should without hesitation think "Let $x=\tan\theta$."

If you're comfortable hyperbolic trig functions, then metacompactness's suggestion is better. :)

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  • $\begingroup$ Weierstrass substitution (wikipedia) $\endgroup$ Commented Jul 4, 2013 at 19:48
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    $\begingroup$ @TheChaz2.0: Say what? The rational parametrization of the circle turns a rational function of $\sin$ and $\cos$ into a rational function of $t$. Why muddy the story with that? This is a poor beginning calculus student!!! $\endgroup$ Commented Jul 4, 2013 at 19:55
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    $\begingroup$ @julien: I am frequently of the opinion that many responders are here to show off and do not think about actual teaching. We can't give a beginning student learning calculus a solution using perverse sheaves, even if it's impressive to a pro. Very frustrating. I feel sorry for franklin.vp's students when he launches into a teaching career. $\endgroup$ Commented Jul 4, 2013 at 20:08
  • $\begingroup$ @TedShifrin I tried this way, but I'm stuck at another integral, again with radix and I don't know how to solve it (see the edit please). The problem is that I don't know a standard criteria for solving them, I neither can rely on my intuition, is there a standard way to solve them? $\endgroup$ Commented Jul 4, 2013 at 20:40
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    $\begingroup$ You need to do enough integrals that experience helps. Note that $\sqrt{1+\tan^2\theta}=\sec\theta$, and this should be on a list of trigonometric integrals that you know or find in your text. In the end, a certain amount of memorization is probably helpful. $\endgroup$ Commented Jul 4, 2013 at 20:48
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your integral can be solved by $x=\sinh t$ and $dx=\cosh t\, dt$ which gives $$\int \frac{\cosh t}{\cosh t} \, dt = \int 1\, dt=t=\sinh^{-1} x = \ln (x+\sqrt{x^2+1})+C,$$ since $\sqrt{1+x^2}=\sqrt{1+\sinh^2 t}=\cosh t$.

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If you are in a mood for miracles, let $w=x+\sqrt{1+x^2}$. Then $$dw= \left(1+\frac{x}{\sqrt{1+x^2}}\right)\,dx=w \frac{dx}{\sqrt{1+x^2}},$$ and we end up needing to find $\displaystyle\int \frac{dw}{w}$.

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    $\begingroup$ This is cute and not hard to guess once one has experience and knows the answer, but I would not give this to a beginning calculus student. $\endgroup$ Commented Jul 4, 2013 at 19:59
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    $\begingroup$ It is a sort of a joke answer, though no more a joke (since equivalent) to the standard first-year calculus miracle integration of $\sec$. I certainly would not give that as an answer in the absence of the other answers. $\endgroup$ Commented Jul 4, 2013 at 20:02
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    $\begingroup$ Yes, @André, I know you well enough. I'm just frustrated by the general tone of the answers and remarks here. I think I'm going to quit answering elementary questions for beginning students. $\endgroup$ Commented Jul 4, 2013 at 20:09
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    $\begingroup$ That's nice, but you could quote your source! $\endgroup$ Commented Jul 4, 2013 at 20:12
  • $\begingroup$ @TedShifrin: if you stop answering elementary questions, who will improve the tone of the answers? It seems like leaving the sheep to the wolves. $\endgroup$ Commented Oct 20, 2014 at 17:18
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This one of the remarkable derivative of the usual hyperbolic function $\mathrm{arsinh}$ and this is the complete list.

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Just to complete the OP's second suggested solution, by the change of variables $x=\tan{\theta}$ we get:

$$\int \frac{dx}{\sqrt{1+x^2}}=\int{\sqrt{1+\tan^2\theta}d\theta}=\int \sqrt{\frac{\sin^2 \theta + \cos^2\theta}{cos^2 \theta}}d\theta=\int \frac{d\theta}{\cos\theta}=\int \sec\theta \ d\theta$$

Now this is a well known integral; to take it, notice that $\frac{d}{d\theta}\left(\tan\theta + \sec \theta \right)=\sec \theta \left(\tan\theta + \sec \theta \right)$. So we will write the integral as:

$$\int \sec \theta \ d\theta = \int \frac{\sec \theta \left(\tan\theta + \sec \theta \right)}{\left(\tan\theta + \sec \theta \right)}d\theta=\ln \left(\sec \theta + \tan\theta \right)$$

Now we have to change back the variables:

$$\int \frac{dx}{\sqrt{1+x^2}}=\ln \left(\sec \left(\tan^{-1}{x} \right) + \tan\left( \tan^{-1}{x} \right) \right)=\ln \left(\sqrt{1+x^2} + x \right)$$

For the last equation, you may want to draw a right-triangle.

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