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

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

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integral of (-2x^4-4x^2-5)

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

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

Solution steps

\int - 2 x^{4} - 4 x^{2} - 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 2 x^{4} d x - \int 4 x^{2} d x - \int 5 d x

\int 2 x^{4} d x = \frac{2 x ^{5}}{5}

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

\int 5 d x = 5 x

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

Add a constant to the solution

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

\int 2 cos (\frac{x}{2}) d x

\frac{\partial}{\partial y} (2 ln (x) + ln (y))

\frac{\partial}{\partial x} (\frac{x}{y})

\int_{- 1}^{1} e^{x^{2}} d x

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

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