What is the integral of (5x-4)^2 ?

The integral of (5x-4)^2 is (25x^3)/3-20x^2+16x+C

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integral of (5x-4)^2

\int (5 x - 4)^{2} d x

\frac{25 x ^{3}}{3} - 20 x^{2} + 16 x + C

Solution steps

\int (5 x - 4)^{2} d x

Expand

(5 x - 4)^{2} : 25 x^{2} - 40 x + 16

= \int 25 x^{2} - 40 x + 16 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 25 x^{2} d x - \int 40 x d x + \int 16 d x

\int 25 x^{2} d x = \frac{25 x ^{3}}{3}

\int 40 x d x = 20 x^{2}

\int 16 d x = 16 x

= \frac{25 x ^{3}}{3} - 20 x^{2} + 16 x

Add a constant to the solution

= \frac{25 x ^{3}}{3} - 20 x^{2} + 16 x + C

f (x) = (3 x + 5)^{3}

f (x) = \frac{5}{x + 2}

y = 4 x^{2}, y = x^{2} + 1

y^{^{'}} + 2 x y = x y^{2}

\frac{\partial}{\partial x} (3 e^{x} sin (y))

What is the integral of (5x-4)^2 ?
The integral of (5x-4)^2 is (25x^3)/3-20x^2+16x+C