What is the integral of x^{1/2}(sqrt(x)-3)^2 ?

The integral of x^{1/2}(sqrt(x)-3)^2 is 2/5 x^{5/2}-3x^2+6x^{3/2}+C

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integral of x^{1/2}(sqrt(x)-3)^2

\int x^{\frac{1}{2}} (x - 3)^{2} d x

\frac{2}{5} x^{\frac{5}{2}} - 3 x^{2} + 6 x^{\frac{3}{2}} + C

Solution steps

\int x^{\frac{1}{2}} (x - 3)^{2} d x

Expand

x^{\frac{1}{2}} (x - 3)^{2} : x^{\frac{3}{2}} - 6 x + 9 x^{\frac{1}{2}}

= \int x^{\frac{3}{2}} - 6 x + 9 x^{\frac{1}{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{3}{2}} d x - \int 6 x d x + \int 9 x^{\frac{1}{2}} d x

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

\int 6 x d x = 3 x^{2}

\int 9 x^{\frac{1}{2}} d x = 6 x^{\frac{3}{2}}

= \frac{2}{5} x^{\frac{5}{2}} - 3 x^{2} + 6 x^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} x^{\frac{5}{2}} - 3 x^{2} + 6 x^{\frac{3}{2}} + C

y^{^{''''}} + 10 y^{^{''}} + 9 y = 0

t \to 0 lim (3 t + 3)

\frac{x - 2}{x - 4}

\int e^{x} cos^{2} (e^{x}) d x

f (x) = x (1 - x)^{4}

What is the integral of x^{1/2}(sqrt(x)-3)^2 ?
The integral of x^{1/2}(sqrt(x)-3)^2 is 2/5 x^{5/2}-3x^2+6x^{3/2}+C