What is the integral of x(6x^2-x+7) ?

The integral of x(6x^2-x+7) is (3x^4)/2-(x^3)/3+(7x^2)/2+C

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integral of x(6x^2-x+7)

\int x (6 x^{2} - x + 7) d x

\frac{3 x ^{4}}{2} - \frac{x ^{3}}{3} + \frac{7 x ^{2}}{2} + C

Solution steps

\int x (6 x^{2} - x + 7) d x

Expand

x (6 x^{2} - x + 7) : 6 x^{3} - x^{2} + 7 x

= \int 6 x^{3} - x^{2} + 7 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 6 x^{3} d x - \int x^{2} d x + \int 7 x d x

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

\int 7 x d x = \frac{7 x ^{2}}{2}

= \frac{3 x ^{4}}{2} - \frac{x ^{3}}{3} + \frac{7 x ^{2}}{2}

Add a constant to the solution

= \frac{3 x ^{4}}{2} - \frac{x ^{3}}{3} + \frac{7 x ^{2}}{2} + C

x, x^{3} - 3 x^{2} + 3 x, - 0, 1

\frac{- 5 x}{2 x - 1}

e^{b x}

\int_{0}^{x} \frac{1}{x ^{2}} d x

\frac{d}{d x} (sin^{4} (3 x))

What is the integral of x(6x^2-x+7) ?
The integral of x(6x^2-x+7) is (3x^4)/2-(x^3)/3+(7x^2)/2+C