Réitigh ∫ x^5+x^3-20x

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

Luacháil

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

Céimeanna Réitigh

Measc an tsuim téarma fá téarma.

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

Fág an leanúnach sna téarmaí as an áireamh.

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

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x^{5}\mathrm{d}x le \frac{x^{6}}{6}.

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

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x^{3}\mathrm{d}x le \frac{x^{4}}{4}.

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

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x\mathrm{d}x le \frac{x^{2}}{2}. Méadaigh -20 faoi \frac{x^{2}}{2}.

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

Má tá F\left(x\right) mar frithdhíorthach do f\left(x\right), beidh tacar do frithdhíorthach uile do f\left(x\right) a thabhairt ag F\left(x\right)+C. Mar sin de, cur an comhtháthú leanúnach C\in \mathrm{R} don toradh.

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

Difreálaigh w.r.t. x

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