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

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

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integral of (6x^8-2/3 x^5+7x^4+sqrt(3))

\int (6 x^{8} - \frac{2}{3} x^{5} + 7 x^{4} + 3) d x

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

Solution steps

\int 6 x^{8} - \frac{2}{3} x^{5} + 7 x^{4} + 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 x^{8} d x - \int \frac{2}{3} x^{5} d x + \int 7 x^{4} d x + \int 3 d x

\int 6 x^{8} d x = \frac{2 x ^{9}}{3}

\int \frac{2}{3} x^{5} d x = \frac{x ^{6}}{9}

\int 7 x^{4} d x = \frac{7 x ^{5}}{5}

\int 3 d x = 3 x

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

Add a constant to the solution

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

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