Réitigh ∫ (x+sqrt[3]{x}+1/x^2) wrt x

$$\int ( x + \sqrt[ 3 ] { x } + \frac { 1 } { x ^ { 2 } } ) d x$$

Luacháil

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}-\frac{1}{x}+С$$

Céimeanna Réitigh

Measc an tsuim téarma fá téarma.

$$\int x\mathrm{d}x+\int \sqrt[3]{x}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x$$

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

$$\frac{x^{2}}{2}+\int \sqrt[3]{x}\mathrm{d}x+\int \frac{1}{x^{2}}\mathrm{d}x$$

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

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}+\int \frac{1}{x^{2}}\mathrm{d}x$$

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}.

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}-\frac{1}{x}$$

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.

$$\frac{x^{2}}{2}+\frac{3x^{\frac{4}{3}}}{4}-\frac{1}{x}+С$$

Difreálaigh w.r.t. x

$$x+\sqrt[3]{x}+\frac{1}{x^{2}}$$