What is the integral of x^{1/2}(-6+6x) ?

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

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integral of x^{1/2}(-6+6x)

\int x^{\frac{1}{2}} (- 6 + 6 x) d x

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

Solution steps

\int x^{\frac{1}{2}} (- 6 + 6 x) d x

x^{\frac{1}{2}} (- 6 + 6 x) = 6 x^{\frac{1}{2}} (- 1 + x)

= \int 6 x^{\frac{1}{2}} (- 1 + x) d x

Take the constant out:

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

= 6 \cdot \int x^{\frac{1}{2}} (- 1 + x) d x

Expand

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

= 6 \cdot \int - x^{\frac{1}{2}} + 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

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

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

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

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

Add a constant to the solution

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

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