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

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

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

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

\frac{8 x ^{5}}{5} + x^{4} - 2 x^{3} - 2 x^{2} + 5 x + C

Solution steps

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

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

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

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

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

\int 5 d x = 5 x

= \frac{8 x ^{5}}{5} + x^{4} - 2 x^{3} - 2 x^{2} + 5 x

Add a constant to the solution

= \frac{8 x ^{5}}{5} + x^{4} - 2 x^{3} - 2 x^{2} + 5 x + C

\frac{d}{d x} (5 x + 2 x^{5})

f (x) = (5 x^{3} + 6 x^{2} + 4)^{8}

y = sin (sin (x)), (2 π, 0)

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + 6 e^{- x}

\int x sin (\frac{1}{10} x) d x

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