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

The integral of (9-x^2)^2 is 81x-6x^3+(x^5)/5+C

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integral of (9-x^2)^2

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

81 x - 6 x^{3} + \frac{x ^{5}}{5} + C

Solution steps

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

Expand

(9 - x^{2})^{2} : 81 - 18 x^{2} + x^{4}

= \int 81 - 18 x^{2} + x^{4} 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 81 d x - \int 18 x^{2} d x + \int x^{4} d x

\int 81 d x = 81 x

\int 18 x^{2} d x = 6 x^{3}

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

= 81 x - 6 x^{3} + \frac{x ^{5}}{5}

Add a constant to the solution

= 81 x - 6 x^{3} + \frac{x ^{5}}{5} + C

\int \frac{2 x}{x ^{2} + 2 x + 2} d x

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

\int u^{- 5} (5 - u^{5} + u^{10}) d u

\int x^{18} d x

What is the integral of (9-x^2)^2 ?
The integral of (9-x^2)^2 is 81x-6x^3+(x^5)/5+C