Լուծեք ∫ x^5+x^3-20x

$$\int x ^ { 5 } + x ^ { 3 } - 20 x$$

Գնահատել

$$\frac{x^{6}}{6}+\frac{x^{4}}{4}-10x^{2}+С$$

Լուծման քայլերը

Integrate the sum term by term.

$$\int x^{5}\mathrm{d}x+\int x^{3}\mathrm{d}x+\int -20x\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{5}\mathrm{d}x+\int x^{3}\mathrm{d}x-20\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{5}\mathrm{d}x with \frac{x^{6}}{6}.

$$\frac{x^{6}}{6}+\int x^{3}\mathrm{d}x-20\int x\mathrm{d}x$$

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

$$\frac{x^{6}}{6}+\frac{x^{4}}{4}-20\int x\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Բազմապատկեք -20 անգամ \frac{x^{2}}{2}:

$$\frac{x^{6}}{6}+\frac{x^{4}}{4}-10x^{2}$$

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.

$$\frac{x^{6}}{6}+\frac{x^{4}}{4}-10x^{2}+С$$

Տարբերակել վերագրած x-ը

$$x\left(x^{4}+x^{2}-20\right)$$