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

The integral of ((x-1)/3)^2 is 1/9 x^3-2/9 x^2+1/9 x-(2x^3}{27}+\frac{x^2)/9+C

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integral of ((x-1)/3)^2

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

\frac{1}{9} x^{3} - \frac{2}{9} x^{2} + \frac{1}{9} x - \frac{2 x ^{3}}{27} + \frac{x ^{2}}{9} + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{2}{9} x (x - 1) d x = \frac{2 x ^{3}}{27} - \frac{x ^{2}}{9}

= \frac{1}{9} x (x - 1)^{2} - (\frac{2 x ^{3}}{27} - \frac{x ^{2}}{9})

Simplify

\frac{1}{9} x (x - 1)^{2} - (\frac{2 x ^{3}}{27} - \frac{x ^{2}}{9}) : \frac{1}{9} x^{3} - \frac{2}{9} x^{2} + \frac{1}{9} x - \frac{2 x ^{3}}{27} + \frac{x ^{2}}{9}

= \frac{1}{9} x^{3} - \frac{2}{9} x^{2} + \frac{1}{9} x - \frac{2 x ^{3}}{27} + \frac{x ^{2}}{9}

Add a constant to the solution

= \frac{1}{9} x^{3} - \frac{2}{9} x^{2} + \frac{1}{9} x - \frac{2 x ^{3}}{27} + \frac{x ^{2}}{9} + C

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

x \to - 3 lim (ln (x^{4} - 3 x + 10))

\int \frac{2 x + 5}{3 x ( x + 3 ) ^{2}} d x

\frac{\partial}{\partial y} (x y + \frac{8}{x} + \frac{8}{y})

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What is the integral of ((x-1)/3)^2 ?
The integral of ((x-1)/3)^2 is 1/9 x^3-2/9 x^2+1/9 x-(2x^3}{27}+\frac{x^2)/9+C