# How to calculate integral of $\int_0^\sqrt[3]4\!\sqrt\frac{x}{4-x^{3/2}}\,\mathrm{d}x$

Given the following integral:

$$\int_0^\sqrt[3]4\!\sqrt\frac{x}{4-x^{3/2}}\,\mathrm{d}x$$

How to solve it? I thought it may be possible to substitute it, but I didn't find anything to substitute. I tried to solve it with Maple, but the CAS didn't get it therefore I don't know how to carry on with this. Can you give me some hints?

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First let $x=u^2$, hopefully after that you will see a final substitution that finishes it off. –  Ragib Zaman Feb 8 '12 at 12:07
It seems like setting $x=u^{2/3}$ would be more effective. Then $\sqrt{x}dx = u^{1/3} \frac{2}{3} u^{-1/3} du = \frac{2}{3} du$, so you are now trying to solve: $$\frac{2}{3}\int_0^2{\frac{du}{\sqrt{4-u}}}$$ –  Thomas Andrews Feb 8 '12 at 14:02

$$\int\sqrt{\frac{x}{4-x^{3/2}}}\,\mathrm dx = -\frac{4\sqrt{x}}{3\sqrt{-\frac{x}{x^{3/2}-4}}}+\mathrm{constant}$$

where you can find the integration steps here by clicking on the button 'Show steps' next to the result.

In your particular case $x\geq 0$ over the whole integration domain such that we may simplify to

$$-\frac{4}{3}\sqrt{4-x^{3/2}}$$

evaluating at $x=0$ and $x=4^{1/3}$ gives the result

$$\int_0^{4^{1/3}}\sqrt{\frac{x}{4-x^{3/2}}}\,\mathrm dx = \frac{4}{3} (2-\sqrt{2})$$

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What happened to the minus sign in the indefinite integral? Shouldn't that simplify to $-\frac 4 3 \sqrt{4-x^{3/2}}$? –  Thomas Andrews Feb 8 '12 at 13:57
Yes, corrected it. –  Till Hoffmann Feb 8 '12 at 14:43

I think it is worth mentioning the case of the integration of the differential binomials.

The expression of the form

$$x^m(a+bx^n)^pdx$$ where $m,n,p,a,b$ are constant is called a differential binomial.

THEOREM. (Piskunov)

The integral

$$\int x^m(a+bx^n)^pdx$$

can be reduced if $m,n,p$ are rational numbers, to the integral of a rational function, and can thus be expressed in terms of elementary functions if:

$1.$ $p$ is an integer.

$2.$ $\dfrac{m+1}{n}$ is an integer.

$3.$ $\dfrac{m+1}{n}+p$ is an integer.

PROOF

We transform the integral writing $x^n = z$ so $dx = \frac 1 n z^{\frac 1 n -1}$. Then:

$$\int {{x^m}} {(a + b{x^n})^p}dx = \int {{z^{{{m + 1} \over n} - 1}}} {(a + bz)^p}dz = \int {{z^q}} {(a + bz)^p}dz$$

$1.$ Let $p$ be an integer. Being $q$ a rational number, let it be $\dfrac r s$. This integral then takes the form $$\int {R\left( {{z^{q/s}},z} \right)dz}$$

which can be reduced by substituting $z=t^s$.

$2.$ If $\dfrac{m+1}{n}$ is an integer. then $q=\dfrac{m+1}{n}-1$ is an integer. $p$ is rational $=\dfrac \lambda \mu$. The integral is reduced to $$\int {R\left( {{z^q},{{\left( {a + bz} \right)}^{{\lambda \over \mu }}}} \right)dz}$$ which can be reduced substituting $a+bz=t^\mu$

$3.$ If $\dfrac{m+1}{n}+p$ is an integer then $\dfrac{m+1}{n}+p-1=q+p$ is an integer. We tranform the integral into

$$\int {{z^{q + p}}{{\left( {{{a + bz} \over z}} \right)}^p}dz}$$

where $q+p$ is an integer and $p=\dfrac \lambda \mu$ is rational. The integral is then

$$\int {R\left[ {z,{{\left( {{{a + bz} \over z}} \right)}^{{\lambda \over \mu }}}} \right]dz}$$

which can be reduced using

$${{a + bz} \over z} = {t^\mu }$$

Note. P.L. Chebyshev, a russian mathematician, proved the integrals just analysed can't be expressed in terms of elementary functions if it isn't the case $1$ , $2$ or $3$.

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