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

The integral of (2x^2-x-1)(3x+4) is (3x^4)/2+(5x^3)/3-(7x^2)/2-4x+C

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

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

\frac{3 x ^{4}}{2} + \frac{5 x ^{3}}{3} - \frac{7 x ^{2}}{2} - 4 x + C

Solution steps

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

Expand

(2 x^{2} - x - 1) (3 x + 4) : 6 x^{3} + 5 x^{2} - 7 x - 4

= \int 6 x^{3} + 5 x^{2} - 7 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^{3} d x + \int 5 x^{2} d x - \int 7 x d x - \int 4 d x

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

= \frac{3 x ^{4}}{2} + \frac{5 x ^{3}}{3} - \frac{7 x ^{2}}{2} - 4 x + C

\frac{d}{d x} (sin^{5} (4 x + 1))

y^{^{'}} \cdot (x^{2} + 2) + 4 x (y^{2} + 2 y + 1) = 0

\int \frac{19}{x ^{3} - 125} d x

\frac{\partial}{\partial x} (3)

f (x) = - 12 x^{\frac{3}{2}} + 7 x^{1.7} + x

What is the integral of (2x^2-x-1)(3x+4) ?
The integral of (2x^2-x-1)(3x+4) is (3x^4)/2+(5x^3)/3-(7x^2)/2-4x+C