# Problem with solving square root integral

I want to ask, I have an example of an integral equation with roots like this :

$$\int \sqrt[3]{x^{2}}\,\operatorname{d}x = \int x^{2/3}\,\operatorname{d}x$$

What if this?

$$\displaystyle\int \sqrt[3]{\sqrt{\sqrt{x^{2}}}}\,\operatorname{d}x = {?}$$

• That's the integral of $x^{2/(2*2*3)}$. – T. Bongers Dec 31 '15 at 9:46

You can easily rewrite your term as

$$\sqrt[3]{\sqrt{\sqrt{x^{2}}}} = (((x^2)^{1/2})^{1/2})^{1/3}=x^{2\cdot (1/2)\cdot(1/2)\cdot(1/3)}=x^{1/6}=\sqrt[6]{x}$$

Be aware that you will lose your $x<0$ part when doing so. If that's the matter, consider

$$\sqrt[3]{\sqrt{\sqrt{x^{2}}}} = \sqrt[3]{\sqrt{|x|}} = |x|^{1/6}$$

And then, split your integral by definition of $|x|$:

$$|x|=\begin{cases}x,&\text{for }x\geqslant 0\\ -x,&\text{for }x<0\end{cases}$$

$$\int \sqrt[3]{\sqrt{\sqrt{x^2}}}\ dx=\int |x|^{\frac12\cdot \frac13}\ dx=\int |x|^{1/6}\ dx$$$$=\begin{cases}\int x^\frac 16dx\text{, if }x\geq0\\ \int (-x)^\frac 16dx\text{, if }x<0\end{cases} = \begin{cases}\dfrac{x^{1+\frac 16}}{1+\frac 16}+C\text{, if }x\geq0\\ -\dfrac{(-x)^{1+\frac 16}}{1+\frac 16}+C\text{, if }x<0\end{cases}$$ $$=\boxed{\dfrac67x\sqrt[6]{|x|}+C}$$