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So I have to do the integration of $\frac{\sqrt{x^2+1}}{x}$. Give me a hint. What should I replace? Should I do it with integration by parts?

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Hint: substitute $x = \tan{y}$; $dx = \sec^2{y} dy$. Then $x^2+1 = \sec^2{y}$.

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You could use the substitution $$ \begin{align} &&u&=x+\sqrt{x^2+1},&x&=\frac{u^2-1}{2u} \\ &&\sqrt{x^2+1}&=\frac{u^2+1}{2u},&dx&=\frac{u^2+1}{2u^2}du \end{align} $$ to get (unless I did some mistake) $$ \int\frac{\sqrt{x^2+1}}{x}dx= \int\frac{\frac{u^2+1}{2u}}{\frac{u^2-1}{2u}}\frac{u^2+1}{2u^2}du= \int\frac{(u^2+1)^2}{2u^2(u^2-1)}du= \int\frac{(u^2-1)^2+4u^2}{2u^2(u^2-1)}du= $$ $$ =\int\left(\frac{(u^2-1)^2}{2u^2(u^2-1)}+\frac{4u^2}{2u^2(u^2-1)}\right)\,du= \int\left(\frac{u^2-1}{2u^2}+\frac{2}{u^2-1}\right)\,du= $$ $$ =\frac{u}{2}+\frac{1}{2u}+\ln|u-1|-\ln|u+1|+C. $$ Now just do the backsubstitution.

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  • $\begingroup$ Interesting substitution! Though, how careful should we be with square root of squares? $\endgroup$ – Math-fun Feb 14 '15 at 9:37
  • $\begingroup$ @Mehdi: I believe it is a well-known substitution, so you should not give me credit for it. Where in this are you concerned about the square root of squares? $\endgroup$ – Mårten W Feb 15 '15 at 22:50
  • $\begingroup$ I just rechecked and see that this is all flawless. I am not sure what was I unhappy about :-)! (I upvoted it yesterday) $\endgroup$ – Math-fun Feb 16 '15 at 9:02
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This problem is a lot uglier than it originally looked. First, let's bring the denominator under the radical.

$$\int\sqrt{\frac{x^2+1}{x^2}}dx=\int\sqrt{1+\frac1{x^2}}dx$$.

Now let's make the substitution rlgordonna suggested.

$$x=\tan y,dx=\sec^2ydy$$ $$\int\sqrt{1+\frac1{\tan^2y}}\sec^2ydy=\int\sqrt{1+\cot^2y}\sec^2dy=\int\csc y\sec^2ydy$$

It will take a little work from here to get this to look like something recognizable.

$$\int\csc y(1+\tan^2y)dy=\int\csc ydy+\int\csc y\tan^2ydy$$

I assume you know how to do the first integral. It's similar to the integral for $\sec y$. As for the second half

$$\int\csc y\tan^2ydy=\int\frac1{\sin y}\times\frac{\sin y}{\cos y}\times\tan ydy=\int\sec y\tan ydy$$

which should now be in a form that looks familiar.

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  • $\begingroup$ Yes, it was quite ugly, but I figured at this level we should avoid suggesting hyperbolics (no offense meant to @Chris'sSister). It should be pointed out that it is urged that OP check his work by differentiating his result; it is actually nice to see the cancellations involved in getting the original function back. $\endgroup$ – Ron Gordon Jan 24 '13 at 22:13

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