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

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

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

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

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

Solution steps

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

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

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

\int C d x = C x

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

Add a constant to the solution

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

2 x^{2} + 2

\frac{3}{s ^{2}}

\int \frac{1}{cos ( 4 x )} d x

x \to 4 lim (\frac{- 6}{4 - x})

\int x^{3} x^{2} + 41 d x

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