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

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

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integral of (x-4)^2x^2

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

\frac{x ^{5}}{5} - 2 x^{4} + \frac{16 x ^{3}}{3} + C

Solution steps

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

Expand

(x - 4)^{2} x^{2} : x^{4} - 8 x^{3} + 16 x^{2}

= \int x^{4} - 8 x^{3} + 16 x^{2} 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 x^{4} d x - \int 8 x^{3} d x + \int 16 x^{2} d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int 8 x^{3} d x = 2 x^{4}

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

= \frac{x ^{5}}{5} - 2 x^{4} + \frac{16 x ^{3}}{3}

Add a constant to the solution

= \frac{x ^{5}}{5} - 2 x^{4} + \frac{16 x ^{3}}{3} + C

\frac{\partial}{\partial x} (- 3 sin (2 x))

\frac{d}{d x} (2 a sin (x))

y^{^{'}} + 4 y = e^{- x}

\int \frac{1 + x ^{2}}{1 + y ^{2}} d x

(x^{2} + 3 y^{2}) d x - 2 x y d y = 0

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