What is the integral of (2x-1)(3x+4) ?

The integral of (2x-1)(3x+4) is 2x^3+(5x^2)/2-4x+C

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integral of (2x-1)(3x+4)

\int (2 x - 1) (3 x + 4) d x

2 x^{3} + \frac{5 x ^{2}}{2} - 4 x + C

Solution steps

\int (2 x - 1) (3 x + 4) d x

Expand

(2 x - 1) (3 x + 4) : 6 x^{2} + 5 x - 4

= \int 6 x^{2} + 5 x - 4 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 6 x^{2} d x + \int 5 x d x - \int 4 d x

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

= 2 x^{3} + \frac{5 x ^{2}}{2} - 4 x + C

\int 0.002 d x

x^{2} - 6 x + 5

y^{^{''}} - 4 y^{^{'}} = 16 t, y (0) = 4, y^{^{'}} (0) = 0

1 - 2 x - 3 x^{2}

x \frac{d y}{d x} = x^{3} + 11 x^{3} y

What is the integral of (2x-1)(3x+4) ?
The integral of (2x-1)(3x+4) is 2x^3+(5x^2)/2-4x+C