What is the integral of 2x^3-2/3 x^2+7x ?

The integral of 2x^3-2/3 x^2+7x is (x^4)/2-(2x^3)/9+(7x^2)/2+C

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integral of 2x^3-2/3 x^2+7x

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

\frac{x ^{4}}{2} - \frac{2 x ^{3}}{9} + \frac{7 x ^{2}}{2} + C

Solution steps

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

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

\int \frac{2}{3} x^{2} d x = \frac{2 x ^{3}}{9}

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

= \frac{x ^{4}}{2} - \frac{2 x ^{3}}{9} + \frac{7 x ^{2}}{2}

Add a constant to the solution

= \frac{x ^{4}}{2} - \frac{2 x ^{3}}{9} + \frac{7 x ^{2}}{2} + C

n = 0 \sum \infty \frac{8}{4 ^{n}}

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

\int t \cdot \frac{1}{e ^{t}} d t

\int (- 2 sec (x) tan (x) - 3 sec^{2} (x)) d x

5 t \cdot \frac{d y}{d t} + 3 y = t

What is the integral of 2x^3-2/3 x^2+7x ?
The integral of 2x^3-2/3 x^2+7x is (x^4)/2-(2x^3)/9+(7x^2)/2+C