What is the integral of ((2x-1)(x+3))/6 ?

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

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integral of ((2x-1)(x+3))/6

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

\frac{1}{6} (\frac{2 x ^{3}}{3} + \frac{5 x ^{2}}{2} - 3 x) + C

Solution steps

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

Take the constant out:

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

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

Expand

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

= \frac{1}{6} \cdot \int 2 x^{2} + 5 x - 3 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}{6} (\int 2 x^{2} d x + \int 5 x d x - \int 3 d x)

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

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

\int 3 d x = 3 x

= \frac{1}{6} (\frac{2 x ^{3}}{3} + \frac{5 x ^{2}}{2} - 3 x)

Add a constant to the solution

= \frac{1}{6} (\frac{2 x ^{3}}{3} + \frac{5 x ^{2}}{2} - 3 x) + C

(0, 0), (2, 1)

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

\frac{\partial}{\partial t} (t e^{- 2 x} + cos (x))

f (x) = x^{2}, at x = 1.9

\int \frac{1 - 0.4 x}{( 1 - x ) ( 5 - x )} d x

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