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

The integral of (3x^2-2x)^3 is (27x^7)/7-9x^6+(36x^5)/5-2x^4+C

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

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

\frac{27 x ^{7}}{7} - 9 x^{6} + \frac{36 x ^{5}}{5} - 2 x^{4} + C

Solution steps

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

Expand

(3 x^{2} - 2 x)^{3} : 27 x^{6} - 54 x^{5} + 36 x^{4} - 8 x^{3}

= \int 27 x^{6} - 54 x^{5} + 36 x^{4} - 8 x^{3} 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 27 x^{6} d x - \int 54 x^{5} d x + \int 36 x^{4} d x - \int 8 x^{3} d x

\int 27 x^{6} d x = \frac{27 x ^{7}}{7}

\int 54 x^{5} d x = 9 x^{6}

\int 36 x^{4} d x = \frac{36 x ^{5}}{5}

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

= \frac{27 x ^{7}}{7} - 9 x^{6} + \frac{36 x ^{5}}{5} - 2 x^{4}

Add a constant to the solution

= \frac{27 x ^{7}}{7} - 9 x^{6} + \frac{36 x ^{5}}{5} - 2 x^{4} + C

\frac{d ^{2} y}{d x ^{2}} = x^{2}

\frac{d}{d x} (4 3 x^{2} + 9)

\int_{0}^{l n (3)} \frac{4 e ^{x}}{1 + e ^{2 x}} d x

x (9^{- 2 x})

y = x + e^{x} - \frac{1}{x}

What is the integral of (3x^2-2x)^3 ?
The integral of (3x^2-2x)^3 is (27x^7)/7-9x^6+(36x^5)/5-2x^4+C