Réitigh ∫ (9/sqrt[4]{t}+4/t^7)dt

$$\int ( \frac { 9 } { \sqrt[ 4 ] { t } } + \frac { 4 } { t ^ { 7 } } ) d t$$

Luacháil

$$12t^{\frac{3}{4}}-\frac{2}{3t^{6}}+С$$

Céimeanna Réitigh

Measc an tsuim téarma fá téarma.

$$\int \frac{9}{\sqrt[4]{t}}\mathrm{d}t+\int \frac{4}{t^{7}}\mathrm{d}t$$

Fág an leanúnach sna téarmaí as an áireamh.

$$9\int \frac{1}{\sqrt[4]{t}}\mathrm{d}t+4\int \frac{1}{t^{7}}\mathrm{d}t$$

Athscríobh \frac{1}{\sqrt[4]{t}} mar t^{-\frac{1}{4}}. Ó \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int t^{-\frac{1}{4}}\mathrm{d}t le \frac{t^{\frac{3}{4}}}{\frac{3}{4}}. Simpligh. Méadaigh 9 faoi \frac{4t^{\frac{3}{4}}}{3}.

$$12t^{\frac{3}{4}}+4\int \frac{1}{t^{7}}\mathrm{d}t$$

Ó \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int \frac{1}{t^{7}}\mathrm{d}t le -\frac{1}{6t^{6}}. Méadaigh 4 faoi -\frac{1}{6t^{6}}.

$$12t^{\frac{3}{4}}-\frac{2}{3t^{6}}$$

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

$$12t^{\frac{3}{4}}-\frac{2}{3t^{6}}+С$$

Difreálaigh w.r.t. t

$$\frac{9}{\sqrt[4]{t}}+\frac{4}{t^{7}}$$