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

The integral of x^{3/2}(4-4x) is 8/5 x^{5/2}-8/7 x^{7/2}+C

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

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

\frac{8}{5} x^{\frac{5}{2}} - \frac{8}{7} x^{\frac{7}{2}} + C

Solution steps

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

Expand

x^{\frac{3}{2}} (4 - 4 x) : 4 x^{\frac{3}{2}} - 4 x^{\frac{5}{2}}

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

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

\int 4 x^{\frac{5}{2}} d x = \frac{8}{7} x^{\frac{7}{2}}

= \frac{8}{5} x^{\frac{5}{2}} - \frac{8}{7} x^{\frac{7}{2}}

Add a constant to the solution

= \frac{8}{5} x^{\frac{5}{2}} - \frac{8}{7} x^{\frac{7}{2}} + C

x \to \infty lim (0.72 x)

\frac{d}{d x} (29 - x^{2} - 4 y^{2})

\frac{d}{d x} (4 cos (x) sin (x))

f (x) = \frac{x ^{3} + 9}{x ^{3}}

\frac{d}{d x} (\frac{x ^{2} + 3 x}{7})

What is the integral of x^{3/2}(4-4x) ?
The integral of x^{3/2}(4-4x) is 8/5 x^{5/2}-8/7 x^{7/2}+C