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

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

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

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

\frac{3 x ^{4}}{4} + \frac{4 x ^{3}}{3} - \frac{7}{9} x + C

Solution steps

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

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

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

\int \frac{7}{9} d x = \frac{7}{9} x

= \frac{3 x ^{4}}{4} + \frac{4 x ^{3}}{3} - \frac{7}{9} x

Add a constant to the solution

= \frac{3 x ^{4}}{4} + \frac{4 x ^{3}}{3} - \frac{7}{9} x + C

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