Réitigh ∫ (6x-6)-(4x^2-16x+12)] wrt x

$$\int ( 6 x - 6 ) - ( 4 x ^ { 2 } - 16 x + 12 ) ] d x$$

Luacháil

$$-\frac{4x^{3}}{3}+11x^{2}-18x+С$$

Céimeanna Réitigh

Chun an mhalairt ar 4x^{2}-16x+12 a aimsiú, aimsigh an mhalairt ar gach téarma.

$$\int 6x-6-4x^{2}+16x-12\mathrm{d}x$$

Comhcheangail 6x agus 16x chun 22x a fháil.

$$\int 22x-6-4x^{2}-12\mathrm{d}x$$

Dealaigh 12 ó -6 chun -18 a fháil.

$$\int 22x-18-4x^{2}\mathrm{d}x$$

Measc an tsuim téarma fá téarma.

$$\int 22x\mathrm{d}x+\int -18\mathrm{d}x+\int -4x^{2}\mathrm{d}x$$

Fág an leanúnach sna téarmaí as an áireamh.

$$22\int x\mathrm{d}x+\int -18\mathrm{d}x-4\int x^{2}\mathrm{d}x$$

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

$$11x^{2}+\int -18\mathrm{d}x-4\int x^{2}\mathrm{d}x$$

Aimsigh suimeálaithe do -18 ag baint úsáid as an tábla do suimeálaithe coitianta riail\int a\mathrm{d}x=ax.

$$11x^{2}-18x-4\int x^{2}\mathrm{d}x$$

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 -4 faoi \frac{x^{3}}{3}.

$$11x^{2}-18x-\frac{4x^{3}}{3}$$

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.

$$11x^{2}-18x-\frac{4x^{3}}{3}+С$$

Difreálaigh w.r.t. x

$$-4x^{2}+22x-18$$