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

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

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

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

\frac{x ^{5}}{5} + \frac{3 x ^{4}}{2} + x^{3} - 9 x^{2} + 9 x + C

Solution steps

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

Expand

(- x^{2} - 3 x + 3)^{2} : x^{4} + 6 x^{3} + 3 x^{2} - 18 x + 9

= \int x^{4} + 6 x^{3} + 3 x^{2} - 18 x + 9 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 6 x^{3} d x + \int 3 x^{2} d x - \int 18 x d x + \int 9 d x

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

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

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

\int 18 x d x = 9 x^{2}

\int 9 d x = 9 x

= \frac{x ^{5}}{5} + \frac{3 x ^{4}}{2} + x^{3} - 9 x^{2} + 9 x

Add a constant to the solution

= \frac{x ^{5}}{5} + \frac{3 x ^{4}}{2} + x^{3} - 9 x^{2} + 9 x + C

\frac{d}{d x} (- \frac{1}{6} x^{3} - \frac{1}{2} x^{2} + x + 1 + k e^{k})

\int 8 sec^{4} (θ) tan^{2} (θ) d θ

\int 3 - 2 x d x

\frac{\partial}{\partial x} (x y + \frac{1}{x} + \frac{8}{y})

n = 1 \sum \infty \frac{1}{( 3 + 2 n ) ^{2}}

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