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

The integral of 6x^4-6x^3+4x^2+C is (6x^5)/5-(3x^4)/2+(4x^3)/3+Cx+C

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

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

\frac{6 x ^{5}}{5} - \frac{3 x ^{4}}{2} + \frac{4 x ^{3}}{3} + C x + C

Solution steps

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

\int 6 x^{4} d x = \frac{6 x ^{5}}{5}

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

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

\int C d x = C x

= \frac{6 x ^{5}}{5} - \frac{3 x ^{4}}{2} + \frac{4 x ^{3}}{3} + C x

Add a constant to the solution

= \frac{6 x ^{5}}{5} - \frac{3 x ^{4}}{2} + \frac{4 x ^{3}}{3} + C x + C

\frac{\partial}{\partial x} (2 x y ln (x))

\frac{\partial}{\partial t} (sin (2 x) cos (2 t))

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

\frac{( 2 s + 12 )}{s ^{2} + 2 s + 5}

\frac{\partial}{\partial x} ((x - 6) (y + 4))

What is the integral of 6x^4-6x^3+4x^2+C ?
The integral of 6x^4-6x^3+4x^2+C is (6x^5)/5-(3x^4)/2+(4x^3)/3+Cx+C