Evaluating $\int_0^{\infty}\frac{\sqrt[3]{x+1}-\sqrt[3]x}{\sqrt x} \operatorname d\!x$ $$\int_0^{\infty}\frac{\sqrt[3]{x+1}-\sqrt[3]x}{\sqrt x}dx$$
I tried with $x=u^6$ then some trigonometric function and other thing but I faild 
what is your suggest to solve ? 
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$\ds{\int_{0}^{\infty}{\root[3]{x + 1}- \root[3]{x} \over \root{x}}\,\dd x:\
    {\Large ?}}$

\begin{align}
&\int_{0}^{\infty}{\root[3]{x + 1}- \root[3]{x} \over \root{x}}\,\dd x=
\int_{0}^{\infty}{1 \over \root{x}}\
\overbrace{%
\bracks{{1 \over 3}\int_{0}^{1}\pars{\mu + x}^{-2/3}\,\dd\mu}}
^{\ds{\root[3]{x + 1}- \root[3]{x}}}\ \dd x
\\[3mm]&={1 \over 3}\int_{0}^{1}\dd\mu\,
\int_{0}^{\infty}{x^{-1/2} \over \pars{\mu + x}^{2/3}}\,\dd x
={1 \over 3}\ \overbrace{\int_{0}^{1}\dd\mu\,\mu^{-1/6}}^{\ds{6 \over 5}}\
\int_{0}^{\infty}{x^{-1/2} \over \pars{1 + x}^{2/3}}\,\dd x
\\[3mm]&={2 \over 5}\int_{0}^{\infty}{x^{-1/2} \over \pars{1 + x}^{2/3}}\,\dd x
\tag{1}
\end{align}

With $\ds{\pars{t \equiv {1 \over 1 + x}\quad\iff\quad x = {1 \over t} - 1}}$,
$\pars{1}$ is reduced to:
\begin{align}
&\color{#00f}{\large%
\int_{0}^{\infty}{\root[3]{x + 1}- \root[3]{x} \over \root{x}}\,\dd x}=
{2 \over 5}\int_{1}^{0}t^{2/3}\pars{1 - t \over t}^{-1/2}\,
\pars{-\,{\dd t \over t^{2}}}
\\[3mm]&={2 \over 5}\int_{0}^{1}t^{-5/6}\pars{1 - t}^{-1/2}\,\dd t
=\color{#00f}{\large{2 \over 5}\,{\rm B}\pars{{1 \over 6},\half}}
\end{align}
where ${\rm B}\pars{x,y}$ is the Beta Function.
A: Is it the exact value of the integral that you are seeking for? Then the answer is
$$ \int_{0}^{\infty} \frac{\sqrt[3]{x+a} - \sqrt[3]{x}}{\sqrt{x}} \, dx = \frac{2}{5}a^{5/6} \beta\left(\frac{1}{6}, \frac{1}{2} \right), \tag{*} $$
where $\beta$ is the beta function. If you are interested in convergence only, you can use the trick
$$ Y - X = \frac{Y^{3} - X^{3}}{Y^{2} + YX + X^{2}}, \quad Y = \sqrt[3]{x+a} \text{ and } X = \sqrt[3]{x}. $$

Solution. Let $I(a)$ denote the integral of (*). Then
$$ I'(a) = \frac{1}{3} \int_{0}^{\infty} \frac{dx}{(x+a)^{2/3}x^{1/2}}. $$
Plugging the substitution $x = a \tan^{2}\theta$, it follows that
$$ I'(a) = \frac{2}{3}a^{-1/6} \int_{0}^{\frac{\pi}{2}} \cos^{-2/3} \theta \,d\theta = \frac{a^{-1/6}}{3} \beta \left( \frac{1}{6}, \frac{1}{3} \right). $$
Since $I(0) = 0$, integrating both sides gives (*).

Justification of some intermediate steps.
We first prove that $I(a)$ is differentiable and Leibniz's integral rule is applicable. Let $a \neq b$. Then
$$ \frac{I(a) - I(b)}{a - b} = \int_{0}^{\infty} \frac{1}{x^{1/3}\{ (x+b)^{2/3} + (x+b)^{1/3}(x+a)^{1/3} + (x+a)^{2/3} \}} \, dx. $$
Let $0 < \delta \leq a, b$ so that in particular both $a$ and $b$ are away from zero. Then the integrand on the right-hand is bounded by the following integrable function:
$$ \frac{1}{3x^{1/2}(x+\delta)^{2/3}}. $$
Thus by dominated convergence theorem, we can let $b \to a$ to obtain the desired result.
Next, by the substitution $x = at$ it follows that $I(a) = a^{5/6}I(1)$. So we can indeed apply the fundamental theorem of calculus as we did in the proof.

Using complex analysis. Consider a keyhole contour $C$ winding $-a$:

If we denote our integrand as $f(x)$, then with the choice of branch-cut of the complex logarithm as $[0, \infty)$, we have
$$ \int_{C} f(z) \, dz = 0. $$
Letting the outer radius $\to \infty$ and the inner radius $\to 0^{+}$, it follows that
$$ \int_{\infty-i0}^{-i0} f(z) \, dz + \int_{-i0}^{-a-i0} f(z) \, dz + \int_{-a+i0}^{i0} f(z) \, dz + \int_{i0}^{\infty+i0} f(z) \, dz = 0, $$
where $i0$ denotes $i$ times a positive infinitesimal, or to be precise, the limit of $i\epsilon$ as $\epsilon \to 0^{+}$. Simplifying, it follows that
$$ I(a)
= \frac{e^{2\pi i/3} + 1}{i (e^{2\pi i/3} + 1)} \int_{0}^{a} \frac{\sqrt[3]{a-x}}{\sqrt{x}} \, dx
= \sqrt{3} \int_{0}^{a} \frac{\sqrt[3]{a-x}}{\sqrt{x}} \, dx. $$
Now with the substitution $x = at$, the right-hand side simplifies in terms of beta function.
