What is the integral of (6+2/5 x^2+3/7 x^3) ?

The integral of (6+2/5 x^2+3/7 x^3) is 6x+(2x^3)/(15)+(3x^4)/(28)+C

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integral of (6+2/5 x^2+3/7 x^3)

\int (6 + \frac{2}{5} x^{2} + \frac{3}{7} x^{3}) d x

6 x + \frac{2 x ^{3}}{15} + \frac{3 x ^{4}}{28} + C

Solution steps

\int 6 + \frac{2}{5} x^{2} + \frac{3}{7} 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 6 d x + \int \frac{2}{5} x^{2} d x + \int \frac{3}{7} x^{3} d x

\int 6 d x = 6 x

\int \frac{2}{5} x^{2} d x = \frac{2 x ^{3}}{15}

\int \frac{3}{7} x^{3} d x = \frac{3 x ^{4}}{28}

= 6 x + \frac{2 x ^{3}}{15} + \frac{3 x ^{4}}{28}

Add a constant to the solution

= 6 x + \frac{2 x ^{3}}{15} + \frac{3 x ^{4}}{28} + C

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What is the integral of (6+2/5 x^2+3/7 x^3) ?
The integral of (6+2/5 x^2+3/7 x^3) is 6x+(2x^3)/(15)+(3x^4)/(28)+C