Trying to solve $\int \sqrt{\sqrt{\sqrt{x}}}\left ( x \frac{1}{x} \right )dx$

Question

I'm trying to solve this integral :

$\int \sqrt{\sqrt{\sqrt{x}}}\left ( x \frac{1}{x} \right )dx = \int x^{\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}}\left ( x\frac{1}{x} \right )dx = \int x^{1/8}\left ( x+x^{-1} \right )dx = \int \left ( x^{9/8}+x^{-1/8} \right )dx = \frac{x^{9/8}}{\frac{9}{8}}+\frac{x^{1/8}}{\frac{1}{8}}+C =\frac{8}{9}x^{9/8}+8x^{1/8}+C $

My answer is correct ?

Answer 1

Notice:

• If it is a multiplication: $$\sqrt{\sqrt{\sqrt{x}}}\left(x\frac{1}{x}\right)=\sqrt{\sqrt{\sqrt{x}}}\left(1\right)=\sqrt{\sqrt{\sqrt{x}}}=x^{\left(\frac{1}{2}\right)^3}=x^{\frac{1}{8}}=\sqrt[8]{x}$$
• If it is a 'mixed fraction': $$x\frac{1}{x}=\frac{x^2+1}{x}\to\sqrt{\sqrt{\sqrt{x}}}\left(x\frac{1}{x}\right)=\sqrt{\sqrt{\sqrt{x}}}\left(\frac{x^2+1}{x}\right)=\sqrt[8]{x^9}+\frac{1}{\sqrt[8]{x^7}}$$

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$$\int\sqrt{\sqrt{\sqrt{x}}}\left(x\frac{1}{x}\right)\space\text{d}x=\int x^{\frac{1}{8}}\space\text{d}x=\frac{8\sqrt[8]{x^9}}{9}+\text{C}$$

$$\int\sqrt{\sqrt{\sqrt{x}}}\left(x\frac{1}{x}\right)\space\text{d}x=\int \left[\sqrt[8]{x^9}+\frac{1}{\sqrt[8]{x^7}}\right]\space\text{d}x=\frac{8}{17}\sqrt[8]{x}\left(x^2+17\right)+\text{C}$$

Answer 2

How did you get from $(x\frac{1}{x})$ to $(x+x^{-1})$? That should just be 1 which turns it into $\int x^{\frac{1}{8}}dx$ which is then trivial.