Solve ∫ (-2x^4-4x^2-5) wrt x

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

Evaluate

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

Solution Steps

Integrate the sum term by term.

$$\int -2x^{4}\mathrm{d}x+\int -4x^{2}\mathrm{d}x+\int -5\mathrm{d}x$$

Factor out the constant in each of the terms.

$$-2\int x^{4}\mathrm{d}x-4\int x^{2}\mathrm{d}x+\int -5\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{4}\mathrm{d}x with \frac{x^{5}}{5}. Multiply -2 times \frac{x^{5}}{5}.

$$-\frac{2x^{5}}{5}-4\int x^{2}\mathrm{d}x+\int -5\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply -4 times \frac{x^{3}}{3}.

$$-\frac{2x^{5}}{5}-\frac{4x^{3}}{3}+\int -5\mathrm{d}x$$

Find the integral of -5 using the table of common integrals rule \int a\mathrm{d}x=ax.

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

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

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

Differentiate w.r.t. x

$$-2x^{4}-4x^{2}-5$$