What is the integral of ((8x^3)/3-3x^2+5) ?

The integral of ((8x^3)/3-3x^2+5) is (2x^4)/3-x^3+5x+C

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integral of ((8x^3)/3-3x^2+5)

\int (\frac{8 x ^{3}}{3} - 3 x^{2} + 5) d x

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

Solution steps

\int \frac{8 x ^{3}}{3} - 3 x^{2} + 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 \frac{8 x ^{3}}{3} d x - \int 3 x^{2} d x + \int 5 d x

\int \frac{8 x ^{3}}{3} d x = \frac{2 x ^{4}}{3}

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

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What is the integral of ((8x^3)/3-3x^2+5) ?
The integral of ((8x^3)/3-3x^2+5) is (2x^4)/3-x^3+5x+C