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

The integral of (5-2/3 x^2+3/4 x^3) is 5x-(2x^3)/9+(3x^4)/(16)+C

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

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

5 x - \frac{2 x ^{3}}{9} + \frac{3 x ^{4}}{16} + C

Solution steps

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

\int 5 d x = 5 x

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

\int \frac{3}{4} x^{3} d x = \frac{3 x ^{4}}{16}

= 5 x - \frac{2 x ^{3}}{9} + \frac{3 x ^{4}}{16}

Add a constant to the solution

= 5 x - \frac{2 x ^{3}}{9} + \frac{3 x ^{4}}{16} + C

x \frac{d y}{d x} + 2 y = 5 x^{3}

\int \frac{39}{39 + e ^{x}} d x

x \to - 2 lim (5 x + 2)

y = x

y^{^{''}} - 2 y^{^{'}} + y = 14.5 e^{t}

What is the integral of (5-2/3 x^2+3/4 x^3) ?
The integral of (5-2/3 x^2+3/4 x^3) is 5x-(2x^3)/9+(3x^4)/(16)+C