Réitigh ∫ x(2+x^2)^2 wrt x

$$\int x ( 2 + x ^ { 2 } ) ^ { 2 } d x$$

Luacháil

$$\frac{x^{6}}{6}+x^{4}+2x^{2}+С$$

Céimeanna Réitigh

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(2+x^{2}\right)^{2} a leathnú.

$$\int x\left(4+4x^{2}+\left(x^{2}\right)^{2}\right)\mathrm{d}x$$

Chun cumhacht a ardú go cumhacht eile, méadaigh na heaspónaint. Iolraigh 2 agus 2 chun 4 a bhaint amach.

$$\int x\left(4+4x^{2}+x^{4}\right)\mathrm{d}x$$

Úsáid an t-airí dáileach chun x a mhéadú faoi 4+4x^{2}+x^{4}.

$$\int 4x+4x^{3}+x^{5}\mathrm{d}x$$

Measc an tsuim téarma fá téarma.

$$\int 4x\mathrm{d}x+\int 4x^{3}\mathrm{d}x+\int x^{5}\mathrm{d}x$$

Fág an leanúnach sna téarmaí as an áireamh.

$$4\int x\mathrm{d}x+4\int x^{3}\mathrm{d}x+\int x^{5}\mathrm{d}x$$

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

$$2x^{2}+4\int x^{3}\mathrm{d}x+\int x^{5}\mathrm{d}x$$

Ó \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} fá choinne k\neq -1, athchuir \int x^{3}\mathrm{d}x le \frac{x^{4}}{4}. Méadaigh 4 faoi \frac{x^{4}}{4}.

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

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

$$2x^{2}+x^{4}+\frac{x^{6}}{6}$$

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.

$$\frac{x^{6}}{6}+x^{4}+2x^{2}+С$$

Difreálaigh w.r.t. x

$$x\left(x^{2}+2\right)^{2}$$