Réitigh ∫ (8x^4-5x^2+10x-5) wrt x

$$\int ( 8 x ^ { 4 } - 5 x ^ { 2 } + 10 x - 5 ) d x$$

Luacháil

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

Céimeanna Réitigh

Measc an tsuim téarma fá téarma.

$$\int 8x^{4}\mathrm{d}x+\int -5x^{2}\mathrm{d}x+\int 10x\mathrm{d}x+\int -5\mathrm{d}x$$

Fág an leanúnach sna téarmaí as an áireamh.

$$8\int x^{4}\mathrm{d}x-5\int x^{2}\mathrm{d}x+10\int x\mathrm{d}x+\int -5\mathrm{d}x$$

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x^{4}\mathrm{d}x le \frac{x^{5}}{5}. Méadaigh 8 faoi \frac{x^{5}}{5}.

$$\frac{8x^{5}}{5}-5\int x^{2}\mathrm{d}x+10\int x\mathrm{d}x+\int -5\mathrm{d}x$$

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x^{2}\mathrm{d}x le \frac{x^{3}}{3}. Méadaigh -5 faoi \frac{x^{3}}{3}.

$$\frac{8x^{5}}{5}-\frac{5x^{3}}{3}+10\int x\mathrm{d}x+\int -5\mathrm{d}x$$

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 10 faoi \frac{x^{2}}{2}.

$$\frac{8x^{5}}{5}-\frac{5x^{3}}{3}+5x^{2}+\int -5\mathrm{d}x$$

Aimsigh suimeálaithe do -5 ag baint úsáid as an tábla do suimeálaithe coitianta riail\int a\mathrm{d}x=ax.

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

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.

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

Difreálaigh w.r.t. x

$$8x^{4}-5\left(x-1\right)^{2}$$