Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given the following integral:


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?

share|improve this question
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
add comment

2 Answers 2

up vote 3 down vote accepted

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

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


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

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

share|improve this answer
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
add comment

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.


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 $3$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.