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

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

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

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

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

Solution steps

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

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

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

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

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

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

\int 3 d x = 3 x

= \frac{4 x ^{6}}{3} - x^{5} - x^{4} - 2 x^{3} - x^{2} - 3 x

Add a constant to the solution

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

\frac{30000}{1.8 t + 9}

\int \frac{1}{4 - 3 x} d x

(2.5) (- 1.7)

x \to 0 lim (\frac{x ^{4}}{x ^{2}})

n = 0 \sum \infty \frac{x ^{n}}{2 ^{n}}

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