Réitigh ∫ (x^5-7+4/x^2) wrt x

$$\int ( x ^ { 5 } - 7 + \frac { 4 } { x ^ { 2 } } ) d x$$

Luacháil

$$\frac{x^{6}}{6}-7x-\frac{4}{x}+С$$

Céimeanna Réitigh

Measc an tsuim téarma fá téarma.

$$\int x^{5}\mathrm{d}x+\int -7\mathrm{d}x+\int \frac{4}{x^{2}}\mathrm{d}x$$

Fág an leanúnach sna téarmaí as an áireamh.

$$\int x^{5}\mathrm{d}x+\int -7\mathrm{d}x+4\int \frac{1}{x^{2}}\mathrm{d}x$$

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

$$\frac{x^{6}}{6}+\int -7\mathrm{d}x+4\int \frac{1}{x^{2}}\mathrm{d}x$$

Aimsigh suimeálaithe do -7 ag baint úsáid as an tábla do suimeálaithe coitianta riail\int a\mathrm{d}x=ax.

$$\frac{x^{6}}{6}-7x+4\int \frac{1}{x^{2}}\mathrm{d}x$$

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int \frac{1}{x^{2}}\mathrm{d}x le -\frac{1}{x}. Méadaigh 4 faoi -\frac{1}{x}.

$$\frac{x^{6}}{6}-7x-\frac{4}{x}$$

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.

$$\frac{x^{6}}{6}-7x-\frac{4}{x}+С$$

Difreálaigh w.r.t. x

$$x^{5}-7+\frac{4}{x^{2}}$$