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

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

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integral of x^2(3-2x)

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

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

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{2 x ^{3}}{3} d x = - \frac{x ^{4}}{6}

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

Simplify

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

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

Add a constant to the solution

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

x^{2} - 6 x, 0

\int_{\frac{π}{4}}^{\frac{π}{3}} cot (x) d x

\int x e^{- 7 x} d x

\frac{\partial}{\partial z} (z e^{x y})

2 + 0.55 sin (\frac{2 π t}{6.2})

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