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

The integral of (4x-3)^2 is (16x^3)/3-12x^2+9x+C

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integral of (4x-3)^2

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

\frac{16 x ^{3}}{3} - 12 x^{2} + 9 x + C

Solution steps

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

Expand

(4 x - 3)^{2} : 16 x^{2} - 24 x + 9

= \int 16 x^{2} - 24 x + 9 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 16 x^{2} d x - \int 24 x d x + \int 9 d x

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

\int 24 x d x = 12 x^{2}

\int 9 d x = 9 x

= \frac{16 x ^{3}}{3} - 12 x^{2} + 9 x

Add a constant to the solution

= \frac{16 x ^{3}}{3} - 12 x^{2} + 9 x + C

x \to 0 + lim ((ln (x))^{x})

\frac{d}{d x} (\frac{1}{x} ln (e^{x} + x))

\frac{\partial}{\partial x} (sin (8 x) cos (2 y))

\int_{1}^{4} x \cdot ln (x) d x

\frac{d}{d x} (\frac{x ^{3}}{3} + \frac{8}{3} - x^{2} + 4)

What is the integral of (4x-3)^2 ?
The integral of (4x-3)^2 is (16x^3)/3-12x^2+9x+C