What is the integral of 5x^4+2/3 x^3+7/3 x^2-16x-4/9 ?

The integral of 5x^4+2/3 x^3+7/3 x^2-16x-4/9 is x^5+(x^4)/6+(7x^3)/9-8x^2-4/9 x+C

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integral of 5x^4+2/3 x^3+7/3 x^2-16x-4/9

\int 5 x^{4} + \frac{2}{3} x^{3} + \frac{7}{3} x^{2} - 16 x - \frac{4}{9} d x

x^{5} + \frac{x ^{4}}{6} + \frac{7 x ^{3}}{9} - 8 x^{2} - \frac{4}{9} x + C

Solution steps

\int 5 x^{4} + \frac{2}{3} x^{3} + \frac{7}{3} x^{2} - 16 x - \frac{4}{9} 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 x^{4} d x + \int \frac{2}{3} x^{3} d x + \int \frac{7}{3} x^{2} d x - \int 16 x d x - \int \frac{4}{9} d x

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

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

\int \frac{7}{3} x^{2} d x = \frac{7 x ^{3}}{9}

\int 16 x d x = 8 x^{2}

\int \frac{4}{9} d x = \frac{4}{9} x

= x^{5} + \frac{x ^{4}}{6} + \frac{7 x ^{3}}{9} - 8 x^{2} - \frac{4}{9} x

Add a constant to the solution

= x^{5} + \frac{x ^{4}}{6} + \frac{7 x ^{3}}{9} - 8 x^{2} - \frac{4}{9} x + C

\int (x^{2} + x + 9) x + 5 d x

\int 2 y - y^{2} d y

\int \frac{2 x - 2}{x ^{2}} d x

\int - e^{3 x - 6} + 1 d x

What is the integral of 5x^4+2/3 x^3+7/3 x^2-16x-4/9 ?
The integral of 5x^4+2/3 x^3+7/3 x^2-16x-4/9 is x^5+(x^4)/6+(7x^3)/9-8x^2-4/9 x+C