What is the integral of (x^2-4x+6) ?

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

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integral of (x^2-4x+6)

\int (x^{2} - 4 x + 6) d x

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

Solution steps

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

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

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

\int 6 d x = 6 x

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

Add a constant to the solution

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

\int \frac{1}{I n x ^{2} ( I n x ) ^{2} - 1} d x

\int \frac{15}{x x ^{2} - 1} d x

sin (x) cos^{2} (x)

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

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