What is the integral of (x^3-3x^2+3x-2) ?

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

integral of (x^3-3x^2+3x-2)

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

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

Solution steps

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

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

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

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

\int 2 d x = 2 x

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

Add a constant to the solution

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

\int (x + sec^{2} (x)) d x

\int 6 sin (x) - 3 cos (x) d x

\int 18 sin^{2} (x) d x

\int 5 x^{3} + e^{x} - cos (x) d x

\int \frac{x ^{2} + 4 x + 2}{x ^{2} + 1} d x

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