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

The integral of (3x^4)/2-4x^{3/2} is (3x^5)/(10)-8/5 x^{5/2}+C

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integral of (3x^4)/2-4x^{3/2}

\int \frac{3 x ^{4}}{2} - 4 x^{\frac{3}{2}} d x

\frac{3 x ^{5}}{10} - \frac{8}{5} x^{\frac{5}{2}} + C

Solution steps

\int \frac{3 x ^{4}}{2} - 4 x^{\frac{3}{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 \frac{3 x ^{4}}{2} d x - \int 4 x^{\frac{3}{2}} d x

\int \frac{3 x ^{4}}{2} d x = \frac{3 x ^{5}}{10}

\int 4 x^{\frac{3}{2}} d x = \frac{8}{5} x^{\frac{5}{2}}

= \frac{3 x ^{5}}{10} - \frac{8}{5} x^{\frac{5}{2}}

Add a constant to the solution

= \frac{3 x ^{5}}{10} - \frac{8}{5} x^{\frac{5}{2}} + C

f (x) = \frac{e ^{x} + e ^{- x}}{2}

\frac{d y}{d x} = \frac{5 x + 2 y}{5 x + 2 y + 2}

\frac{d ^{2} y}{d x ^{2}} = 0

11 x - 4 y x^{2} + 1 y^{^{'}} = 0, y (0) = 3

\frac{d y}{d x} = \frac{y + x cos ^{2} ( \frac{y}{x} )}{x}

What is the integral of (3x^4)/2-4x^{3/2} ?
The integral of (3x^4)/2-4x^{3/2} is (3x^5)/(10)-8/5 x^{5/2}+C