What is the integral of sqrt(x)(x^2-4x) ?

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

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integral of sqrt(x)(x^2-4x)

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

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

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

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

\int e^{5 x} x^{2} d x

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

\int t^{2} e^{- t} d t

x \to 2 lim (\frac{1}{x ^{2} - 1})

y = \frac{1}{8 x ^{2}}, y = x, y = \frac{x}{8}

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