What is the integral of (x^2-9)^4 ?

The integral of (x^2-9)^4 is (x^9)/9-(36x^7)/7+(486x^5)/5-972x^3+6561x+C

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integral of (x^2-9)^4

\int (x^{2} - 9)^{4} d x

\frac{x ^{9}}{9} - \frac{36 x ^{7}}{7} + \frac{486 x ^{5}}{5} - 972 x^{3} + 6561 x + C

Solution steps

\int (x^{2} - 9)^{4} d x

Expand

(x^{2} - 9)^{4} : x^{8} - 36 x^{6} + 486 x^{4} - 2916 x^{2} + 6561

= \int x^{8} - 36 x^{6} + 486 x^{4} - 2916 x^{2} + 6561 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^{8} d x - \int 36 x^{6} d x + \int 486 x^{4} d x - \int 2916 x^{2} d x + \int 6561 d x

\int x^{8} d x = \frac{x ^{9}}{9}

\int 36 x^{6} d x = \frac{36 x ^{7}}{7}

\int 486 x^{4} d x = \frac{486 x ^{5}}{5}

\int 2916 x^{2} d x = 972 x^{3}

\int 6561 d x = 6561 x

= \frac{x ^{9}}{9} - \frac{36 x ^{7}}{7} + \frac{486 x ^{5}}{5} - 972 x^{3} + 6561 x

Add a constant to the solution

= \frac{x ^{9}}{9} - \frac{36 x ^{7}}{7} + \frac{486 x ^{5}}{5} - 972 x^{3} + 6561 x + C

\int e^{x} a^{x} d x

x^{2} y^{^{'}} + y = \frac{x}{y}, y (1) = 4

f (x) = x + 2, - 2, 1

\frac{( y ^{^{'}} - e - t + 4 )}{y} = - 4

x \to 0 lim (\frac{7 ^{s i n (x)} - 1}{x})

What is the integral of (x^2-9)^4 ?
The integral of (x^2-9)^4 is (x^9)/9-(36x^7)/7+(486x^5)/5-972x^3+6561x+C