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

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

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

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

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

Solution steps

\int \frac{x ^{3} + 4 x ^{2} - x}{x} d x

Simplify

\frac{x ^{3} + 4 x ^{2} - x}{x} : x^{2} + 4 x - 1

= \int x^{2} + 4 x - 1 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^{2} d x + \int 4 x d x - \int 1 d x

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

\int 4 x d x = 2 x^{2}

\int 1 d x = x

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

Add a constant to the solution

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

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