What is the integral of (x^2-4x+5) ?

The integral of (x^2-4x+5) is (x^3)/3-2x^2+5x+C

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integral of (x^2-4x+5)

\int (x^{2} - 4 x + 5) d x

\frac{x ^{3}}{3} - 2 x^{2} + 5 x + C

Solution steps

\int x^{2} - 4 x + 5 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 5 d x

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

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

\int 5 d x = 5 x

= \frac{x ^{3}}{3} - 2 x^{2} + 5 x

Add a constant to the solution

= \frac{x ^{3}}{3} - 2 x^{2} + 5 x + C

\frac{d}{d x} (arcsin (\frac{x}{10}))

\frac{\partial}{\partial x} (- \frac{4}{x ^{2} y})

d er i v a t i v e f (t) = t^{\frac{1}{2}} (13 t + 2)

d er i v a t i v e \frac{9}{( 9 - x ^{8} ) ^{\frac{3}{2}}}

\int_{3}^{5} \frac{x ^{2}}{8} d x

What is the integral of (x^2-4x+5) ?
The integral of (x^2-4x+5) is (x^3)/3-2x^2+5x+C