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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

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\int x \cdot (x + 5)^{4} d x

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\int (3 x^{2} + 1)^{3} 6 x d x

\int \frac{a ^{2}}{x x ^{2} - a ^{2}} d x

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