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

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

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

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

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

Solution steps

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

\int 8 x^{5} d x = \frac{4 x ^{6}}{3}

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

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

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