What is the integral of-x^4+((9x^2))/4-2x+3 ?

The integral of-x^4+((9x^2))/4-2x+3 is -(x^5)/5+(3x^3)/4-x^2+3x+C

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integral of-x^4+((9x^2))/4-2x+3

\int - x^{4} + \frac{( 9 x ^{2} )}{4} - 2 x + 3 d x

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

Solution steps

\int - x^{4} + \frac{9 x ^{2}}{4} - 2 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^{4} d x + \int \frac{9 x ^{2}}{4} d x - \int 2 x d x + \int 3 d x

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

\int \frac{9 x ^{2}}{4} d x = \frac{3 x ^{3}}{4}

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

(2 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0, y (1) = 5

\frac{\partial}{\partial x} ((1 - y) + 10 (x - y^{2})^{2})

x y^{^{'}} + 3 y = x^{2} - 1

\frac{\partial}{\partial x} (\frac{x}{y} + \frac{y}{x})

(x + a) e^{3 x} + b

What is the integral of-x^4+((9x^2))/4-2x+3 ?
The integral of-x^4+((9x^2))/4-2x+3 is -(x^5)/5+(3x^3)/4-x^2+3x+C