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I tried a lot of different approaches but found this problem very hard.

So can you help me with this integral?

$$\int {\sqrt{\frac{x+1}{x}}}\:dx$$

Thanks.

Olivier Oloa
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user107157
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3 Answers3

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HINT:

Substitute $u=\frac{1+x}{x}$ and $\text{d}u=\left(\frac{1}{x}-\frac{1+x}{x^2}\right)\space\text{d}x$:

$$\mathcal{I}\left(x\right)=\int\sqrt{\frac{1+x}{x}}\space\text{d}x=-\int\frac{\sqrt{u}}{\left(1-u\right)^2}\space\text{d}u$$

Now,substitute $s=\sqrt{u}$ and $\text{d}s=\frac{1}{2\sqrt{u}}\space\text{d}u$:

$$-\int\frac{\sqrt{u}}{\left(1-u\right)^2}\space\text{d}u=-2\int\frac{s^2}{\left(s^2-1\right)^2}\space\text{d}s$$

Now, use partial frations:

$$\frac{s^2}{\left(s^2-1\right)^2}=\frac{1}{4(s-1)^2}+\frac{1}{4(s-1)}+\frac{1}{4(1+s)^2}-\frac{1}{4(1+s)}$$

Jan Eerland
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Hint

If $u=\sqrt{\frac{x+1}{x}}$, your integral becomes $$-\int\frac{2u}{(u^2-1)^3}\mathrm d u.$$

Surb
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  • Hi. I'm slightly dim-witted because I live on Jupiter so bear with me. Also maths on Jupiter is very strange. But if your equation was correct then I could substitute $v=u^{2}-1$ and get $dv = 2u du$ and hence the equation becomes $-\int{\frac{1}{v^{3}}}$ which would be $\frac{v^{2}}{2}$ and simplified to $\frac{(u^{2}-1)^{2}}{2}$ and $\frac{1}{2x^{2}}$. But seriously I'm not that good at maths. Maybe Earthlings can teach me. – user107157 Nov 06 '16 at 11:10
  • Additionally $\frac{d}{dx} \frac{1}{2x^{2}}$ is not $\sqrt{\frac{x+1}{x}}$. Am I missing something. – user107157 Nov 06 '16 at 11:24
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    @user107157: Yes ! – Surb Nov 06 '16 at 11:40
  • I created an answer with the solution that I think you were hinting at. +1 to you – user107157 Nov 08 '16 at 22:03
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I have finally understood what Surb meant in his answer and I'm posting it here for the benefit of future readers. I can't post it as a comment due to space restrictions.

$\int{\sqrt{\frac{x+1}{x}}}$

Let u = $\sqrt{\frac{x+1}{x}}$ and v = $\frac{x+1}{x} = 1 + \frac{1}{x}$

$\therefore u = \sqrt{v}$

$\frac{dv}{dx} = \frac{-1}{x^{2}}$

$\frac{du}{dx} = \frac{du}{dv} \frac{dv}{dx} = \frac{1}{2 \sqrt{v}} \frac{-1}{x^{2}} = \frac{-1}{2ux^{2}}$

$u^{2} = 1 + \frac{1}{x} \therefore u^{2} - 1 = \frac{1}{x} \therefore x^{2} = \frac{1}{( u^{2} - 1)^{2}}$

$\therefore \frac{du}{dx} = \frac{- ( u^{2} - 1)^{2}}{2u}$

$\therefore dx = \frac{-2u}{ ( u^{2} - 1)^{2}} du$

$\therefore \int{\sqrt{\frac{x+1}{x}}}dx = \int{ \frac{-2u^{2}}{ ( u^{2} - 1)^{2}}}du = -2 \int{ \frac{u^{2}}{ ( u^{2} - 1)^{2}}}du = -2 \int{ u \times \frac{u}{ ( u^{2} - 1)^{2}}}du$

Using rule $\int{a b'} = ab - \int{b a'}$ where a = u and b' = $\frac{u}{ ( u^{2} - 1)^{2}}$

Let v = $u^{2} - 1$ then

b = $\int{b'} = \int{ \frac{u}{ ( u^{2} - 1)^{2}}}du = \frac{1}{2} \int{\frac{1}{v^{2}}}dv = \frac{-1}{2(u^{2} - 1)}$

$\int{a b'} = u \times \frac{-1}{2(u^{2} - 1)} - \int{ \frac{-1}{2(u^{2} - 1)}}du = \frac{-u}{2(u^{2} - 1)} + \frac{1}{2} \int{\frac{1}{u^{2} - 1}}du$

$\frac{1}{u^{2} - 1} = \frac{1}{u+1} \frac{1}{u-1} = \frac{1}{2(u-1)} - \frac{1}{2(u+1)}$

$\therefore \int{\frac{1}{u^{2} - 1}}du = \int{\frac{1}{2(u-1)} - \frac{1}{2(u+1)}}du = \frac{log(u-1) - log(u+1)}{2}$

$\therefore \int{a b'} = \frac{-u}{2(u^{2} - 1)} + \frac{log(u-1) - log(u+1)}{4}$

$\therefore \int{\sqrt{\frac{x+1}{x}}}dx = \frac{u}{u^{2} - 1} + \frac{log(u+1) - log(u-1)}{2} = x \sqrt{\frac{x+1}{x}} + \frac{log( \sqrt{\frac{x+1}{x}}+1 ) - log( \sqrt{\frac{x+1}{x}} - 1)}{2}$

Drumroll please. And thanks Surb for the hint. You are awesome. You really helped Jupiter.

user107157
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