What is the integral of (x^4-9x)^2(4x^3-9) ?

The integral of (x^4-9x)^2(4x^3-9) is (x^{12})/3-9x^9+81x^6-243x^3+C

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integral of (x^4-9x)^2(4x^3-9)

\int (x^{4} - 9 x)^{2} (4 x^{3} - 9) d x

\frac{x ^{12}}{3} - 9 x^{9} + 81 x^{6} - 243 x^{3} + C

Solution steps

\int (x^{4} - 9 x)^{2} (4 x^{3} - 9) d x

Expand

(x^{4} - 9 x)^{2} (4 x^{3} - 9) : 4 x^{11} - 81 x^{8} + 486 x^{5} - 729 x^{2}

= \int 4 x^{11} - 81 x^{8} + 486 x^{5} - 729 x^{2} 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 4 x^{11} d x - \int 81 x^{8} d x + \int 486 x^{5} d x - \int 729 x^{2} d x

\int 4 x^{11} d x = \frac{x ^{12}}{3}

\int 81 x^{8} d x = 9 x^{9}

\int 486 x^{5} d x = 81 x^{6}

\int 729 x^{2} d x = 243 x^{3}

= \frac{x ^{12}}{3} - 9 x^{9} + 81 x^{6} - 243 x^{3}

Add a constant to the solution

= \frac{x ^{12}}{3} - 9 x^{9} + 81 x^{6} - 243 x^{3} + C

x \to 0 lim (\frac{cot ( π x ) \cdot sin ( x )}{2 sec ( x )})

\int sin (x) cos (k x) d x

\int 9 (9 x + 4) d x

f (x) = x^{2} - 2 cos (x) - 3

x y^{2} y^{^{'}} = y^{3} - x^{3}

What is the integral of (x^4-9x)^2(4x^3-9) ?
The integral of (x^4-9x)^2(4x^3-9) is (x^{12})/3-9x^9+81x^6-243x^3+C