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So,once I asked to know the integral $$\int \frac{\sqrt{x^2+1}}{x}dx$$ ... and the advice I got was to subsitute $x=\tan u$.

How about substituting $\sqrt{x^2+1}=u$? Will it work that way?

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Well, what happens if you do? You don't expect us to do all the work for you? –  Harald Hanche-Olsen Jan 30 '13 at 19:09
    
Chill out,lost "Olsen" twin.It was just a question. –  user60290 Jan 30 '13 at 19:10
    
Well, you will generally find that people are more interested in answering your questions if it is clear that you have put some work into them before just asking. You'll get higher quality answers, too. Not that there is anything low quality about the answers you got here. (And is my name that funny? I didn't pick it myself.) –  Harald Hanche-Olsen Jan 31 '13 at 15:41
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3 Answers 3

Yes, that will work. Rewrite our integral as $$\int \frac{x\sqrt{x^2+1}}{x^2}\,dx.$$

Make your substitution. I prefer to write it as $x^2+1=u^2$. So $2x\,dx=2u\,du$. We end up with $$\int \frac{u^2}{u^2-1} \, du.$$ This will be familiar if you have already covered partial fractions.

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so the result with be smth like sqrt(x^2+1) +1/2*[sqrt(x^2+1)-1]/[sqrt(x^2+1)+1]? –  user60290 Jan 30 '13 at 19:24
    
Our integrand turns out to be $1+\frac{1}{u^2-1}=1+\frac{1}{2}\left(\frac{1}{u-1}-\frac{1}{u+1}\right)$. When you integrate you will get logarithms. This is routine partial fractions, so I assume you have not yet covered that method. –  André Nicolas Jan 30 '13 at 19:33
    
I have ,sweet french man,thank you anyway :) –  user60290 Jan 30 '13 at 19:37
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Hint: If you are familiar to hyperbolic trigonometric functions, try to substitute $$x=\sinh(t)$$ We know that $\cosh^2(t)-\sinh^2(t)=1$

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Nice hint, without just spitting out an answer! +1 :-) –  amWhy Jan 31 '13 at 0:12
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Yes, it works. It's not pretty, but it works.

$$\int \frac{\sqrt{x^2 + 1}}{x} dx$$ $$u = \sqrt{x^2 + 1}\implies x = \pm \sqrt{u^2-1}$$ $$du = -(x^2 + 1)^{-\frac{1}{2}}2x dx \implies dx = -(\sqrt{x^2+1})\frac{1}{2x}du$$ $$dx = -\left(\sqrt{u^2 - 1+1}\right)\frac{1}{\pm2\sqrt{u^2-1}}du = \mp\frac{u}{2\sqrt{u^2-1}}du$$

Substitute into the integral: $$\int \frac{u}{\pm \sqrt{u^2-1}}\left(\mp\frac{u}{2\sqrt{u^2-1}}\right)du$$ $$\int \frac{u^2}{u^2-1}du$$

$$\int 1 + \frac{1}{(u-1)(u+1)}du$$ Note that, by partial fractions, $$\frac{1}{(u-1)(u+1)} = \frac{1}{2(u-1)} - \frac{1}{2(u+1)}$$

Thus, you have the integral: $$\int 1 + \frac{1}{2(u-1)} - \frac{1}{2(u+1)}du$$

Evaluating: $$u + \frac{1}{2}\ln(u-1) - \frac{1}{2}\ln(u+1) + C$$ Simplify: $$u + \frac{1}{2}\ln\left(\frac{u-1}{u+1}\right) + C$$

Now substitute back in: $$\sqrt{x^2 + 1} + \frac{1}{2}\ln\left(\frac{\sqrt{x^2 + 1}-1}{\sqrt{x^2 + 1}+1}\right) + C$$ $$\sqrt{x^2 + 1} + \frac{1}{2}\ln\left(\frac{x^2 + 1 - 1}{\left(\sqrt{x^2 + 1}+1\right)^2}\right) + C$$ $$\sqrt{x^2 + 1} + \frac{1}{2}\ln(x^2) - \frac{1}{2}\ln\left({\left(\sqrt{x^2 + 1}+1\right)^2}\right) + C$$

Yielding the solution:

$$\sqrt{x^2 + 1} + \ln(x) - \ln\left({\sqrt{x^2 + 1}+1}\right) + C$$

Verification from W|A.

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