# Calculation of a integral

$$\int \left(1 - \frac{1}{x^2}\right) \sqrt{x \sqrt{x}} \, dx$$

Could anyone help me calculate this integral? Thanks in advance.

• $$\sqrt{x\sqrt x}=(x\cdot x^{\dfrac12})^{\dfrac12}=\left(x^{1+\dfrac12}\right)^{\dfrac12}=x^{\dfrac34}$$ – lab bhattacharjee Apr 5 '14 at 16:48

Your integral is $$\int\left(1-\frac1{x^2}\right)\sqrt{x\sqrt{x}}\,{\rm d}x$$

$$=\int (x^{3/4} - x^{-5/4})\,{\rm d}x$$

$$\frac{4x^{7/4}}{7} +4x^{-1/4}+C$$

Because

$$\sqrt{x\sqrt{x}} =\sqrt{x}\sqrt[4]{x} = x^{1/2}x^{1/4}=x^{3/4}$$

• Got a parenthesis error in your first line. – Thomas Andrews Apr 5 '14 at 16:52
• @ThomasAndrews thanks. – Guy Apr 5 '14 at 16:53
• One of the answers has -1/4 on 4x, and you have 1/4, which one is the correct one? One question as well, how do we get x^(3/4) from 1-1/x^2? – user133022 Apr 5 '14 at 16:54
• @user133022 Multiply $x^{3/4}\times1=x^{3/4}$ and $x^{-2}\times x^{-3/4}=x^{-5/4}$. The typo is mine. Correcting. – Guy Apr 5 '14 at 16:56

$$\int x^{3/4}dx-\int x^{-5/4}dx=\dfrac{4x^{7/4}}{7}+4x^{-1/4}+C$$