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

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

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integral of (3x^3-2x^2+6x)/x

\int \frac{3 x ^{3} - 2 x ^{2} + 6 x}{x} d x

x^{3} - x^{2} + 6 x + C

Solution steps

\int \frac{3 x ^{3} - 2 x ^{2} + 6 x}{x} d x

Expand

\frac{3 x ^{3} - 2 x ^{2} + 6 x}{x} : \frac{3 x ^{3}}{x} - \frac{2 x ^{2}}{x} + \frac{6 x}{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 \frac{3 x ^{3}}{x} d x - \int \frac{2 x ^{2}}{x} d x + \int \frac{6 x}{x} d x

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

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

\int \frac{6 x}{x} d x = 6 x

= x^{3} - x^{2} + 6 x

Add a constant to the solution

= x^{3} - x^{2} + 6 x + C

\int e^{x} cos (y) - x - 2

(sech (2 x))^{s i n (3 x)}

\int csc (x) d x

1 + cos^{2} (x)

\frac{d x}{d y} 2 - \frac{d y}{d x} sec (x) = 0

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