What is the integral of 6x^4-6x^3+5x^2+C ?

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

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integral of 6x^4-6x^3+5x^2+C

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

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

Solution steps

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

\int C d x = C x

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

Add a constant to the solution

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

\frac{50}{1 - 3 s ^{2}}

\frac{\partial}{\partial x} (s^{8} t + s t^{- 6})

5 x y^{5} + 8 x y = 39, (3, 1)

\int_{1}^{3} \frac{3}{x ^{2} + 4 x + 4} d x

sin (θ) + θ cos (θ)

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