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

The integral of x^4-x^2-x is (x^5)/5-(x^3)/3-(x^2)/2+C

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integral of x^4-x^2-x

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

\frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + C

Solution steps

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

\int x^{4} d x = \frac{x ^{5}}{5}

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

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

= \frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{5}}{5} - \frac{x ^{3}}{3} - \frac{x ^{2}}{2} + C

\int 6 x^{2} x + 2 d x

\int \frac{x ^{3} + 11}{x ^{2} + 5 x + 6} d x

\frac{\partial}{\partial x} (4 x^{3} y^{2} - 2)

d er i v a t i v e y = x^{4} - 2 x^{3} - 9

\int (x - 1) d x

What is the integral of x^4-x^2-x ?
The integral of x^4-x^2-x is (x^5)/5-(x^3)/3-(x^2)/2+C