What is the integral of (6x^5-12x^4+9x^3)/(3x^2) ?

The integral of (6x^5-12x^4+9x^3)/(3x^2) is 1/6 (3x^4-8x^3+9x^2)+C

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integral of (6x^5-12x^4+9x^3)/(3x^2)

\int \frac{6 x ^{5} - 12 x ^{4} + 9 x ^{3}}{3 x ^{2}} d x

\frac{1}{6} (3 x^{4} - 8 x^{3} + 9 x^{2}) + C

Solution steps

\int \frac{6 x ^{5} - 12 x ^{4} + 9 x ^{3}}{3 x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{6 x ^{5} - 12 x ^{4} + 9 x ^{3}}{x ^{2}} d x

Expand

\frac{6 x ^{5} - 12 x ^{4} + 9 x ^{3}}{x ^{2}} : \frac{6 x ^{5}}{x ^{2}} - \frac{12 x ^{4}}{x ^{2}} + \frac{9 x ^{3}}{x ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int \frac{6 x ^{5}}{x ^{2}} d x - \int \frac{12 x ^{4}}{x ^{2}} d x + \int \frac{9 x ^{3}}{x ^{2}} d x)

\int \frac{6 x ^{5}}{x ^{2}} d x = \frac{3 x ^{4}}{2}

\int \frac{12 x ^{4}}{x ^{2}} d x = 4 x^{3}

\int \frac{9 x ^{3}}{x ^{2}} d x = \frac{9 x ^{2}}{2}

= \frac{1}{3} (\frac{3 x ^{4}}{2} - 4 x^{3} + \frac{9 x ^{2}}{2})

Simplify

\frac{1}{3} (\frac{3 x ^{4}}{2} - 4 x^{3} + \frac{9 x ^{2}}{2}) : \frac{1}{6} (3 x^{4} - 8 x^{3} + 9 x^{2})

= \frac{1}{6} (3 x^{4} - 8 x^{3} + 9 x^{2})

Add a constant to the solution

= \frac{1}{6} (3 x^{4} - 8 x^{3} + 9 x^{2}) + C

\frac{d y}{d x} = 2 y x + y x^{2}

\int 7 θ cos (θ) d θ

\frac{d}{d x} (x (4 - x^{2}))

- x \frac{d y}{d x} = x^{2} - y

x \to 0 lim (\frac{3}{x ( 2 \cdot x + 1 )})

What is the integral of (6x^5-12x^4+9x^3)/(3x^2) ?
The integral of (6x^5-12x^4+9x^3)/(3x^2) is 1/6 (3x^4-8x^3+9x^2)+C