What is the integral of (x^3+6x^2-4x+5) ?

The integral of (x^3+6x^2-4x+5) is (x^4)/4+2x^3-2x^2+5x+C

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integral of (x^3+6x^2-4x+5)

\int (x^{3} + 6 x^{2} - 4 x + 5) d x

\frac{x ^{4}}{4} + 2 x^{3} - 2 x^{2} + 5 x + C

Solution steps

\int x^{3} + 6 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^{3} d x + \int 6 x^{2} d x - \int 4 x d x + \int 5 d x

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

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

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

\int 5 d x = 5 x

= \frac{x ^{4}}{4} + 2 x^{3} - 2 x^{2} + 5 x

Add a constant to the solution

= \frac{x ^{4}}{4} + 2 x^{3} - 2 x^{2} + 5 x + C

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\int sin (5 y) d y

\int e^{x} 10 + e^{x} d x

What is the integral of (x^3+6x^2-4x+5) ?
The integral of (x^3+6x^2-4x+5) is (x^4)/4+2x^3-2x^2+5x+C