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How can I calculate this $$\int_1^{+\infty} \frac {dx}{x^3\sqrt{x^2+x}}$$

I have no idea what to do with it, integration by parts or by substitution doesn't work for me.

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We have, as $x \to 0^+$, $$ \frac 1{x^3\sqrt{x^2+x}}\sim \frac 1{x^{7/2}} $$ and since $7/2>1$, the integral $\displaystyle \int_0^{+\infty} \frac {dx}{x^3\sqrt{x^2+x}}$ is not convergent.


Since the OP has changed the initial integral, a route for the new one is to make the change of variable $u=\dfrac1x$.

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Whoops, I made a typo, I meant 1 instead of 0. Sorry – Involutive Automorphism Jan 10 at 15:03

Sub $x \mapsto 1/u$...

$$\int_1^{\infty} \frac{dx}{x^3 \sqrt{x^2+x}} = \int_0^1 du \frac{u^2}{\sqrt{1+u}} = \int_1^2 dv \, v^{-1/2} (v-1)^2 = \int_1^2 dv \, \left (v^{3/2} - 2 v^{1/2} + v^{-1/2} \right )$$

which is

$$\frac25 \left (2^{5/2} - 1 \right ) -2 \frac23 \left (2^{3/2} - 1 \right ) + 2 \left (2^{1/2}-1 \right ) = \frac{14}{15} \sqrt{2} - \frac{16}{15}$$

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Can you please state clearly what $u$ and $v$ are? – Involutive Automorphism Jan 10 at 15:21
    
$u=1/x$, $v=u+1$. – Ron Gordon Jan 10 at 15:22
    
How to come up with such an idea? :) – Involutive Automorphism Jan 10 at 16:43

First of all, I've checked my answer with Mathematice and I'm right.

$$\int_{1}^{\infty}\frac{1}{x^3\sqrt{x^2+x}}\space\text{d}x=$$ $$\lim_{n\to\infty}\int_{1}^{n}\frac{1}{x^3\sqrt{x^2+x}}\space\text{d}x=$$ $$\lim_{n\to\infty}\int_{1}^{n}\frac{1}{x^3\sqrt{\left(x+\frac{1}{2}\right)^2-\frac{1}{4}}}\space\text{d}x=$$


Substitute $u=x+\frac{1}{2}$ and $\text{d}u=\text{d}x$.

This gives a new lower bound $u=1+\frac{1}{2}=\frac{3}{2}$ and upper bound $u=n+\frac{1}{2}$:


$$\lim_{n\to\infty}\int_{\frac{3}{2}}^{n+\frac{1}{2}}\frac{1}{\left(u-\frac{1}{2}\right)^3\sqrt{u^2-\frac{1}{4}}}\space\text{d}u=$$


Substitute $s=\text{arcsec}(2u)$ and $\text{d}u=\frac{\tan(s)\sec(s)}{2}\space\text{d}s$.

This gives a new lower bound $s=\text{arcsec}\left(2\cdot\frac{3}{2}\right)=\text{arcsec}(3)$ and

upper bound $s=\text{arcsec}\left(2\cdot\left(n+\frac{1}{2}\right)\right)=\text{arcsec}\left(2n+1\right)$:


$$\frac{1}{2}\lim_{n\to\infty}\int_{\text{arcsec}(3)}^{\text{arcsec}\left(2n+1\right)}\frac{2\sec(s)}{\left(\frac{\sec(s)}{2}-\frac{1}{2}\right)^3}\space\text{d}s=$$ $$\lim_{n\to\infty}\int_{\text{arcsec}(3)}^{\text{arcsec}\left(2n+1\right)}\frac{\sec(s)}{\left(\frac{\sec(s)}{2}-\frac{1}{2}\right)^3}\space\text{d}s=$$


Substitute $p=\tan\left(\frac{s}{2}\right)$ and $\text{d}p=\frac{\sec^2\left(\frac{s}{2}\right)}{2}\space\text{d}s$.

This gives a new lower bound $p=\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)$ and upper bound $p=\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)$:


$$\lim_{n\to\infty}\int_{\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)}^{\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)}\frac{(p^2-1)^2}{p^6}\space\text{d}p=$$ $$2\lim_{n\to\infty}\int_{\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)}^{\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)}\left[\frac{1}{p^6}-\frac{2}{p^4}+\frac{1}{p^2}\right]\space\text{d}p=$$ $$2\lim_{n\to\infty}\left[\left[-\frac{1}{5p^5}\right]_{\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)}^{\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)}-2\left[-\frac{1}{3p^3}\right]_{\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)}^{\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)}+\left[-\frac{1}{p}\right]_{\tan\left(\frac{\text{arcsec}\left(3\right)}{2}\right)}^{\tan\left(\frac{\text{arcsec}\left(2n+1\right)}{2}\right)}\right]=$$ $$\frac{14\sqrt{2}}{15}-\lim_{x\to\infty}\space\frac{2\sqrt{\frac{1+n}{1+2n}}\sqrt{1+2n}\left(3-4n+8n^2\right)}{15\sqrt{n^5}}=$$


  • Apply L'Hôpitals rule;
  • Use the product rule;
  • Use the power rule;

$$\frac{14\sqrt{2}}{15}-\frac{16}{15}$$

So:

$$\int_{1}^{\infty}\frac{1}{x^3\sqrt{x^2+x}}\space\text{d}x=\frac{14\sqrt{2}}{15}-\frac{16}{15}$$

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How are both other answers, posted several hours before, not vastly superior? – Did Jan 12 at 7:26

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