What is the integral of (x-9)^2 ?

The integral of (x-9)^2 is (x^3)/3-9x^2+81x+C

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integral of (x-9)^2

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

\frac{x ^{3}}{3} - 9 x^{2} + 81 x + C

Solution steps

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

Expand

(x - 9)^{2} : x^{2} - 18 x + 81

= \int x^{2} - 18 x + 81 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^{2} d x - \int 18 x d x + \int 81 d x

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

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

\int 81 d x = 81 x

= \frac{x ^{3}}{3} - 9 x^{2} + 81 x

Add a constant to the solution

= \frac{x ^{3}}{3} - 9 x^{2} + 81 x + C

\int_{0}^{\frac{π}{3}} tan^{5} (x) d x

y^{^{''}} + \frac{6}{x} y^{^{'}} + \frac{4}{x ^{2}} y = 0

f (x) = x^{2} - 4 x + 2

x \to 1 lim (e^{(x^{3} - x)})

\frac{d}{d x} (e^{π x})