What is the integral of (t^{3/2}+2t^{1/2}-4t^{-1/2}) ?

The integral of (t^{3/2}+2t^{1/2}-4t^{-1/2}) is 2/5 t^{5/2}+4/3 t^{3/2}-8t^{1/2}+C

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integral of (t^{3/2}+2t^{1/2}-4t^{-1/2})

\int (t^{\frac{3}{2}} + 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{2}}) d t

\frac{2}{5} t^{\frac{5}{2}} + \frac{4}{3} t^{\frac{3}{2}} - 8 t^{\frac{1}{2}} + C

Solution steps

\int t^{\frac{3}{2}} + 2 t^{\frac{1}{2}} - 4 t^{- \frac{1}{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 t^{\frac{3}{2}} d t + \int 2 t^{\frac{1}{2}} d t - \int 4 t^{- \frac{1}{2}} d t

\int t^{\frac{3}{2}} d t = \frac{2}{5} t^{\frac{5}{2}}

\int 2 t^{\frac{1}{2}} d t = \frac{4}{3} t^{\frac{3}{2}}

\int 4 t^{- \frac{1}{2}} d t = 8 t^{\frac{1}{2}}

= \frac{2}{5} t^{\frac{5}{2}} + \frac{4}{3} t^{\frac{3}{2}} - 8 t^{\frac{1}{2}}

Add a constant to the solution

= \frac{2}{5} t^{\frac{5}{2}} + \frac{4}{3} t^{\frac{3}{2}} - 8 t^{\frac{1}{2}} + C

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What is the integral of (t^{3/2}+2t^{1/2}-4t^{-1/2}) ?
The integral of (t^{3/2}+2t^{1/2}-4t^{-1/2}) is 2/5 t^{5/2}+4/3 t^{3/2}-8t^{1/2}+C