What is the integral of x^3-4x^2+x+6 ?

The integral of x^3-4x^2+x+6 is (x^4)/4-(4x^3)/3+(x^2)/2+6x+C

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integral of x^3-4x^2+x+6

\int x^{3} - 4 x^{2} + x + 6 d x

\frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2} + 6 x + C

Solution steps

\int x^{3} - 4 x^{2} + x + 6 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 6 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}

\int 6 d x = 6 x

= \frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2} + 6 x

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + \frac{x ^{2}}{2} + 6 x + C

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What is the integral of x^3-4x^2+x+6 ?
The integral of x^3-4x^2+x+6 is (x^4)/4-(4x^3)/3+(x^2)/2+6x+C