How to evaluate $\int{\frac{\sqrt{x}}{1 + \sqrt[3]{x}}}dx$?

Question

$$\int{\frac{\sqrt{x}}{1 + \sqrt[3]{x}}}dx$$

When I first came across the above integral I noticed it is unlikely an immediate $u$-sub as it involves both square and cube roots. So, the regular substitutions for evaluating $\sqrt{f(x)}$ like trigonometric/hyperbolic will not work with $\sqrt[3]{f(x)}$.

Therefore I tried to change the form first to $\displaystyle\int{\frac{1}{\frac1{\sqrt{x}} + \sqrt[6]{x}}}dx$ or use the identity $1 + a^3 = (1 + a)(a^2 + a + 1)$ with $a = \sqrt[3]{x}$ to transform to $\displaystyle\int{\frac{\sqrt{x}\cdot( \sqrt[3]{x^2} + \sqrt[3]{x} + 1)}{1 + x}}dx$.

The latter only becomes worse and the former does not open ways to clear substitutions.

But, there is an answer to the integral exists, as evaluated by my CAS:

$$\int{\frac{\sqrt{x}}{1 + \sqrt[3]{x}}}dx = \frac{6x^{7/6}}{7} - \frac{6x^{5/6}}{5} + 2\sqrt{x} - 6x^{1/6} + \arctan{\left(x^{1/6}\right)} + C$$.

It is surprising as there are terms like $\arctan$ when the integral does not have the form $\frac{1}{1 + \dots^2}$ in it or yet.

Hence I am posting this question.

Answer 1

Since the HCF of $\frac12$ and $\frac13$ is $\frac16$, the substitution $t=\sqrt[6]{x}$ helps in simplifying the integral:

$$6\int\frac{t^8}{1+t^2}\mathrm dt$$ $$=6\int t^6-t^4+t^2-1+\frac1{1+t^2}\mathrm dt$$

You can take it from here.

Answer 2

Why the substitution of $\sqrt[6]{x}=t$?

Note : LCM = Least common multiple

$$I=\displaystyle\int{\frac{\sqrt{x}}{1 + \sqrt[3]{x}}}dx$$ Re-writing the above integral as,

$$I=\displaystyle\int x^{\frac{1}{2}}(1+x^\frac{1}{3})^{-1}dx$$

Here's a generalized approach $$I=\int x^m(a+bx^n)^p\,dx$$

Case (1):

If $p$ is an integer, for $p>0$, apply the binomial theorem and for $p<0$, take LCM of denominators of $m$ and $n$, substitute $x=t^N$, where $N$ is LCM of $m,n$.

Case (2):

If $\frac{m+1}{n}$ is an integer, substitute $a+bx^n=t^N$, where $N$ is denominator of $p$.

Case (3):

If $\frac{m+1}{n}+p$ is an integer, substitute $ax^{-1}+b=t^N$, where $N$ is denominator of $p$.

Your integral matches with the second sub case of case$(1)$; taking LCM of denominators of $m$ and $n$,

The LCM of the denominators of m and n is LCM(2,3)=6.

Go ahead with the substitution $x=t^6\implies \sqrt[6]x=t$ and you have three answers that explain further.

Answer 3

The substitution $u=\sqrt[6]x$ leads to $$I=6\int \frac{u^8}{1+u^2}\mathrm du $$ which can be solved using partial fraction decomposition.

Answer 4

Too long for a comment

Considering the more general problem of $$I_n=\int \frac{x^{\frac{1}{n}}}{1+x^{\frac{1}{n+1}}}\,dx$$ which has a solution in terms of the Gaussian hypergeometric function.

Using, as in other answers, $x=t^{n(n+1)}$ $$I_n=n(n+1)\int \frac{t^{n (n+2)}}{1+t^n}\,dt$$ Using long division, we have simple expressions. For example, $$I_4=\int \frac{x^{\frac{1}{4}}}{1+x^{\frac{1}{5}}}\,dx=20\int \frac{ t^{24}}{1+t^4}\,dt$$

$$\frac{ t^{24}}{1+t^4}=-1+t^4-t^8+t^{12}-t^{16}+t^{20}+\frac{1}{1+t^4}$$ and then $$I_4=-20 t+4 t^5-\frac{20}{9} t^9+\frac{20 }{13}t^{13}-\frac{20 }{17}t^{17}+ \frac{20}{21}t^{21}+$$ $$5 \sqrt{2}\tan ^{-1}\left(\frac{t\sqrt{2} }{1-t^2}\right)+\frac{5}{\sqrt{2}}\log \left(\frac{t^2+\sqrt{2} t+1}{t^2-\sqrt{2} t+1}\right)$$

All the problem resides in the last terms which do not show major difficulties playing with the roots of unity.

In general, $$\frac{t^{n (n+2)}}{1+t^n}=\sum_{k=0}^{n+1} (-1)^{k+1}\,t^{kn}+ R_n$$

Edit

For those who enjoy hypergeometric functions $$I_n=\frac{n}{n+1}\,x^{\frac{n+1}{n}}\,\,\, _2F_1\left(1,n+\frac{1}{n}+2;n+\frac{1}{n}+3;-x^{\frac{1}{n+1}}\right)$$

Answer 5

Your integral is pretty simple and tricky.

I want to do it in a different way.

My Process :

Let $$I=\int\frac{\sqrt{x}}{1+\sqrt[3]{x}}dx$$

Now my trick is write $\sqrt[3]{x}=(\sqrt{x})^{\frac{2}{3}}$

Therefore the integral is becoming $$I=\int\frac{\sqrt{x}}{1+(\sqrt{x})^{\frac{2}{3}}}dx$$

Now, substitute $\sqrt{x}=t$

Therefore $\frac{1}{2\sqrt{x}}dx=dt$

$\implies dx=2\sqrt{x}dt=2tdt$

$$\implies I=2\int\frac{t^{2}}{1+t^{\frac{2}{3}}}dt$$

Now, substitute $t=\tan^{3}(u)$

Therefore finally you will get the integral to be $$6\int\tan^{8}(u)du$$

Now, the next steps are very easy.

Write $\tan^{8}(u)=\tan^{6}(u)\tan^{2}(u)=\tan^{6}(u)\sec^{2}(u)-\tan^{6}(u)$

Similarly proceed for $\tan^{6}(u)$ in the same way.

Answer 6

Notice that $$I=\int\frac{1}{6x^{5/6 }}\cdot\frac{6x^{4/3}}{\sqrt[3]x+1}dx$$

Under $u = \sqrt[6]{x}\to \mathrm{d}u = \frac{1}{6x^{5/6}} \, \mathrm{d}x$

$$I=\int \frac{u^{8}}{u^{2} + 1} \, \mathrm{d}u$$

By polynomial long division $\frac{u^{8}}{u^{2} + 1}= \left(\frac{1}{u^{2} + 1} + u^{6} - u^{4} + u^{2} - 1\right)$ , therefore

$$= \int \left(\frac{1}{u^{2} + 1} + u^{6} - u^{4} + u^{2} - 1\right) \mathrm{d}u= \frac{6x^{\frac{7}{6}}}{7} - \frac{6x^{\frac{5}{6}}}{5} + 2 \sqrt{x} - 6 \sqrt[6]{x} + 6 \arctan\left(\sqrt[6]{x}\right) + C$$

Which can be simplified to exactly the answer you got.