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

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

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

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

\frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - \frac{x ^{2}}{2} + C

Solution steps

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

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

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

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

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{4 x ^{3}}{3} - \frac{x ^{2}}{2} + C

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