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

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

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

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

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

Solution steps

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

\int 3 x^{4} d x = \frac{3 x ^{5}}{5}

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

\int 8 tan (x) sin (y) cos (y) d x

\frac{d}{d x} ((2 x^{2} - 7)^{5} - (1 + 4 x^{3})^{5})

\frac{3 x}{e ^{x}}

(x \cdot e^{x})^{^{'}}

\frac{d}{d x} (- x cos (3 x))

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