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Trying to solve: $$\int\sqrt{\frac{x+1}{x}}dx$$ I was thinking solving with substition: $$t=\frac{x+1}{x}$$ which would lead to $$dx=-\frac{1}{(t+1)^2}dt$$ The result being $$-\int\frac{\sqrt t}{(t+1)^2}dt$$
How do I proceed from here? Or better yet, is there an easier way?

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$$x = \sinh^2 t$$

$$\text{d}x = 2\cosh(t)\sinh(t)\ \text{d}t$$

$$x+1 = \cosh^2(t)$$

Hence

$$2\int \sqrt{\frac{\cosh^2 t}{\sinh^2 t}} \cosh t\sinh t\ \text{d}t$$

$$2\int \cosh^2 t\ \text{d}t$$

Which is trivial:

$$2\left(\frac{1}{2}\left(t + \cosh t\sinh t\right)\right)$$

So

$$t + \cosh t\sinh t$$

Now you are surely able to come back to $x$.

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  • $\begingroup$ can you show how you got $dx=2cosh(t)sinh(t)dt$? $\endgroup$ – user2974951 Nov 20 '16 at 12:34
  • $\begingroup$ @user2974951 It's about Hyperbolic Functions! Search on Wikipedia, it's quite similar to the Sine and Cosine way. $\endgroup$ – Von Neumann Nov 20 '16 at 12:37
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What about substituting

$$t^2=\frac{x+1}x=1+\frac1x\iff x=\frac1{t^2-1} \implies 2tdt=-\frac{dx}{x^2}\implies$$

$$dx=-\frac{2t\,dt}{(t^2-1)^2}\implies \int\sqrt\frac{x+1}x\;dx=-\int \frac{2t^2}{(t^2-1)^2}dt$$

and the above can be done more or less simply with partial fractions:

$$\frac{2t^2}{(t-1)^2(t+1)^2}=\frac A{t-1}+\frac B{(t-1)^2}+\frac C{t+1}+\frac D{(t+1)^2}\implies$$

$$2t^2=A(t-1)(t+1)^2+B(t+1)^2+C(t-1)^2(t+1)+D(t-1)^2$$

and etc.

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    $\begingroup$ This is really a tedious way to proceed. $\endgroup$ – Von Neumann Nov 20 '16 at 12:26
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    $\begingroup$ @AlanTuring Perhaps, but it gets the work done...:) $\endgroup$ – DonAntonio Nov 20 '16 at 12:31
  • $\begingroup$ Haha that is true too :D $\endgroup$ – Von Neumann Nov 20 '16 at 12:32
  • $\begingroup$ Sometimes one see this way, and that one sees another way. Yours is faster and less messier...but it requires to know hyperbolic functions and some hyperbolic identities. It is just a matter of luck...and taste, of course. $\endgroup$ – DonAntonio Nov 20 '16 at 12:33
  • $\begingroup$ You're right! I mean your answer may be "slow" but it works great. Indeed I don't understand the down vote on your answer too -.- $\endgroup$ – Von Neumann Nov 20 '16 at 12:36
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Rationalise the numerator:

$\displaystyle \int \sqrt{\frac{x+1}{x}} \, \text{d}x = \int \frac{x+1}{\sqrt{x^2+x}} \, \text{d}x = \int \frac{2x+1}{2\sqrt{x^2+x}} + \frac{1}{2\sqrt{x^2+x}} \, \text{d}x$

The first part is a simple application of the reverse chain rule. The second part is the inverse hyperbolic sine formula, which is trivial once the square is completed.

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\begin{align} Let \sqrt{x+1}+\sqrt{x}=t,\sqrt{x+1}-\sqrt{x}=1/t,\sqrt{x+1}=(t+1/t)/2,\sqrt{x}=(t-1/t)/2\\ x=(t-1/t)^2/4,dx=(t-1/t)(1+1/t^2)/2dt,\\ I=\frac{1}{2}\int\left(t+\frac{2}{t}+\frac{1}{t^3}\right)dt\\ =\frac{1}{2}\left(\frac{t^2}{2}+2\ln |t|-\frac{1}{2t^2}\right)+C\\ =\sqrt{x+1}\sqrt{x}+\ln |\sqrt{x+1}+\sqrt{x}|+C \end{align}

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By the change of variable $2x=t-1$,

$$2\int\sqrt{\frac {x+1}x}dx=\int \sqrt{\frac{t+1}{t-1}}dt=\int\frac{t+1}{\sqrt{t^2-1}}dt=\sqrt{t^2-1}+\text{arsinh } t.$$

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