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

The integral of ((2-x)/2)^2 is 1/4 x^3-(x^3)/6-1/2 x^2+x+C

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

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

\frac{1}{4} x^{3} - \frac{x ^{3}}{6} - \frac{1}{2} x^{2} + x + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{4} x (2 - x)^{2} - \int - \frac{1}{2} x (2 - x) d x

= \frac{1}{4} x (2 - x)^{2} - \int - \frac{1}{2} x (2 - x) d x

\int - \frac{1}{2} x (2 - x) d x = - \frac{1}{2} (x^{2} - \frac{x ^{3}}{3})

= \frac{1}{4} x (2 - x)^{2} - (- \frac{1}{2} (x^{2} - \frac{x ^{3}}{3}))

Simplify

\frac{1}{4} x (2 - x)^{2} - (- \frac{1}{2} (x^{2} - \frac{x ^{3}}{3})) : \frac{1}{4} x^{3} - \frac{x ^{3}}{6} - \frac{1}{2} x^{2} + x

= \frac{1}{4} x^{3} - \frac{x ^{3}}{6} - \frac{1}{2} x^{2} + x

Add a constant to the solution

= \frac{1}{4} x^{3} - \frac{x ^{3}}{6} - \frac{1}{2} x^{2} + x + C

6 t

\frac{d}{d x} (\frac{4}{1 - x})

y^{^{'}} = 13 y - y^{2} - 40

\int \frac{1}{x 9 x ^{2} - 1} d x

x^{^{'}} = \frac{1}{1 + t ^{2}}

What is the integral of ((2-x)/2)^2 ?
The integral of ((2-x)/2)^2 is 1/4 x^3-(x^3)/6-1/2 x^2+x+C