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

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

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

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

\frac{2 x ^{3}}{3} + \frac{x ^{2}}{2} - 3 x + C

Solution steps

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

Simplify

\frac{2 x ^{3} + x ^{2} - 3 x}{x} : 2 x^{2} + x - 3

= \int 2 x^{2} + x - 3 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 2 x^{2} d x + \int x d x - \int 3 d x

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

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

\int 3 d x = 3 x

= \frac{2 x ^{3}}{3} + \frac{x ^{2}}{2} - 3 x

Add a constant to the solution

= \frac{2 x ^{3}}{3} + \frac{x ^{2}}{2} - 3 x + C

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