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

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

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integral of (x^2-4x+3)

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

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

Solution steps

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

\int sin^{4} (2 x) cos^{3} (2 x) d x

\frac{d y}{d x} = \frac{( x y + y ^{2} + x ^{2} )}{x ^{2}}

\frac{\partial}{\partial x} ((3 x + 4 y) e^{y})

\int_{1}^{x} t y + 3 d t

\int_{1}^{2} x^{5} ln (x) d x

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