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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

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

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

Add a constant to the solution

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

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