What is the integral of (x^3-6x^2+9x+6) ?

The integral of (x^3-6x^2+9x+6) is (x^4)/4-2x^3+(9x^2)/2+6x+C

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integral of (x^3-6x^2+9x+6)

\int (x^{3} - 6 x^{2} + 9 x + 6) d x

\frac{x ^{4}}{4} - 2 x^{3} + \frac{9 x ^{2}}{2} + 6 x + C

Solution steps

\int x^{3} - 6 x^{2} + 9 x + 6 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 6 x^{2} d x + \int 9 x d x + \int 6 d x

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

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

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

\int 6 d x = 6 x

= \frac{x ^{4}}{4} - 2 x^{3} + \frac{9 x ^{2}}{2} + 6 x

Add a constant to the solution

= \frac{x ^{4}}{4} - 2 x^{3} + \frac{9 x ^{2}}{2} + 6 x + C

\int \frac{34}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\int \frac{5 x ^{2} - 3 x + 2}{( x - 1 ) ( x ^{2} + 1 )} d x

\int (3 x^{\frac{1}{2}} + x^{4}) d x

\int 7 sec^{3} (x) d x

\int 9 x tan^{2} (3 x) d x

What is the integral of (x^3-6x^2+9x+6) ?
The integral of (x^3-6x^2+9x+6) is (x^4)/4-2x^3+(9x^2)/2+6x+C