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

The integral of x^4(x^2-3x)^2 is (x^9)/9-(3x^8)/4+(9x^7)/7+C

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

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

\frac{x ^{9}}{9} - \frac{3 x ^{8}}{4} + \frac{9 x ^{7}}{7} + C

Solution steps

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

Expand

x^{4} (x^{2} - 3 x)^{2} : x^{8} - 6 x^{7} + 9 x^{6}

= \int x^{8} - 6 x^{7} + 9 x^{6} 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 6 x^{7} d x + \int 9 x^{6} d x

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

\int 6 x^{7} d x = \frac{3 x ^{8}}{4}

\int 9 x^{6} d x = \frac{9 x ^{7}}{7}

= \frac{x ^{9}}{9} - \frac{3 x ^{8}}{4} + \frac{9 x ^{7}}{7}

Add a constant to the solution

= \frac{x ^{9}}{9} - \frac{3 x ^{8}}{4} + \frac{9 x ^{7}}{7} + C

y^{^{''}} + 4 y = tan (2 t)

\int \frac{- sin ( x )}{cos ( x )} d x

\int \frac{2 x}{x ^{2} - 4} d x

\int 2 e^{- 2 x} sin (2 x) d x

\frac{d y}{d x} - 2 y tan (x) = y^{2} tan^{2} (x)

What is the integral of x^4(x^2-3x)^2 ?
The integral of x^4(x^2-3x)^2 is (x^9)/9-(3x^8)/4+(9x^7)/7+C