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

The integral of (sqrt(x))/2 (x-1)^2 is 1/105 x^{3/2}(15x^2-42x+35)+C

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integral of (sqrt(x))/2 (x-1)^2

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

\frac{1}{105} x^{\frac{3}{2}} (15 x^{2} - 42 x + 35) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

x (x - 1)^{2} : x^{\frac{5}{2}} - 2 x^{\frac{3}{2}} + x

= \frac{1}{2} \cdot \int x^{\frac{5}{2}} - 2 x^{\frac{3}{2}} + x 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

= \frac{1}{2} (\int x^{\frac{5}{2}} d x - \int 2 x^{\frac{3}{2}} d x + \int x d x)

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

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

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

= \frac{1}{2} (\frac{2}{7} x^{\frac{7}{2}} - \frac{4}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}})

Simplify

\frac{1}{2} (\frac{2}{7} x^{\frac{7}{2}} - \frac{4}{5} x^{\frac{5}{2}} + \frac{2}{3} x^{\frac{3}{2}}) : \frac{1}{105} x^{\frac{3}{2}} (15 x^{2} - 42 x + 35)

= \frac{1}{105} x^{\frac{3}{2}} (15 x^{2} - 42 x + 35)

Add a constant to the solution

= \frac{1}{105} x^{\frac{3}{2}} (15 x^{2} - 42 x + 35) + C

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

x \to + 0 + lim (x^{\frac{2}{x}})

y^{^{''}} + 15 y^{^{'}} + 54 y = 0

f (x) = (x + 2)^{\frac{2}{3}}

\int \frac{1}{2} (1 + cos (2 x)) d x

What is the integral of (sqrt(x))/2 (x-1)^2 ?
The integral of (sqrt(x))/2 (x-1)^2 is 1/105 x^{3/2}(15x^2-42x+35)+C