What is the integral of (2x^5-x^2)^2(x^{(-1)/2}) ?

The integral of (2x^5-x^2)^2(x^{(-1)/2}) is 8/21 x^{21/2}-8/15 x^{15/2}+2/9 x^{9/2}+C

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integral of (2x^5-x^2)^2(x^{(-1)/2})

\int (2 x^{5} - x^{2})^{2} (x^{\frac{- 1}{2}}) d x

\frac{8}{21} x^{\frac{21}{2}} - \frac{8}{15} x^{\frac{15}{2}} + \frac{2}{9} x^{\frac{9}{2}} + C

Solution steps

\int (2 x^{5} - x^{2})^{2} x^{\frac{- 1}{2}} d x

\frac{- 1}{2} = - \frac{1}{2}

= \int (2 x^{5} - x^{2})^{2} x^{- \frac{1}{2}} d x

Expand

(2 x^{5} - x^{2})^{2} x^{- \frac{1}{2}} : 4 x^{\frac{19}{2}} - 4 x^{\frac{13}{2}} + x^{\frac{7}{2}}

= \int 4 x^{\frac{19}{2}} - 4 x^{\frac{13}{2}} + x^{\frac{7}{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 4 x^{\frac{19}{2}} d x - \int 4 x^{\frac{13}{2}} d x + \int x^{\frac{7}{2}} d x

\int 4 x^{\frac{19}{2}} d x = \frac{8}{21} x^{\frac{21}{2}}

\int 4 x^{\frac{13}{2}} d x = \frac{8}{15} x^{\frac{15}{2}}

\int x^{\frac{7}{2}} d x = \frac{2}{9} x^{\frac{9}{2}}

= \frac{8}{21} x^{\frac{21}{2}} - \frac{8}{15} x^{\frac{15}{2}} + \frac{2}{9} x^{\frac{9}{2}}

Add a constant to the solution

= \frac{8}{21} x^{\frac{21}{2}} - \frac{8}{15} x^{\frac{15}{2}} + \frac{2}{9} x^{\frac{9}{2}} + C

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What is the integral of (2x^5-x^2)^2(x^{(-1)/2}) ?
The integral of (2x^5-x^2)^2(x^{(-1)/2}) is 8/21 x^{21/2}-8/15 x^{15/2}+2/9 x^{9/2}+C