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

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

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

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

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

Solution steps

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

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

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

\int 8 d x = 8 x

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

Add a constant to the solution

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

2 x \frac{d y}{d x} + 2 y = x^{3} y^{4}

\int 18 y^{2} d y

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

\int_{2}^{4} \frac{x + 2}{x ^{2} + 3 x - 4} d x

\frac{d}{d x} (- \frac{1}{4 x ^{3}})

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