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

The integral of (4x-3)(2x-1) is (8x^3)/3-5x^2+3x+C

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

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

\frac{8 x ^{3}}{3} - 5 x^{2} + 3 x + C

Solution steps

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

Expand

(4 x - 3) (2 x - 1) : 8 x^{2} - 10 x + 3

= \int 8 x^{2} - 10 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

= \int 8 x^{2} d x - \int 10 x d x + \int 3 d x

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 10 x d x = 5 x^{2}

\int 3 d x = 3 x

= \frac{8 x ^{3}}{3} - 5 x^{2} + 3 x

Add a constant to the solution

= \frac{8 x ^{3}}{3} - 5 x^{2} + 3 x + C

\frac{\partial}{\partial x} (x^{3} y + x y^{2} - 2 x^{2} - y^{2})

\int \frac{x - 1}{x + 2} d x

\int u \cdot sec^{2} (u) d u

f (x) = 8 x^{3}

\frac{d}{d x} ((sec^{2} (x)) d x)

What is the integral of (4x-3)(2x-1) ?
The integral of (4x-3)(2x-1) is (8x^3)/3-5x^2+3x+C