Whakaoti i te ∫ (from 1 to 3) of (x^2+1/x^2+1) wrt x

$$\int _ { 1 } ^ { 3 } ( x ^ { 2 } + \frac { 1 } { x ^ { 2 } } + 1 ) d x$$

Aromātai

$$\frac{34}{3}\approx 11.333333333$$

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kōmitimititia te kīanga tapeke mā te kīanga.

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

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int x^{2}\mathrm{d}x ki te \frac{x^{3}}{3}.

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

Nā te mea \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} mō te k\neq -1, me whakakapi \int \frac{1}{x^{2}}\mathrm{d}x ki te -\frac{1}{x}.

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

Kimihia te tau tōpū o 1 mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

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

Ko te tau tōpū tautuhi ko te pārōnaki kōaro o te kīanga i aromātaitia i te tepe tōrunga o te pāwhaitua, tangohia te pārōnaki kōaro i aromātaitia i te tepe tōraro o te pāwhaitua.

$$\frac{3^{3}}{3}-3^{-1}+3-\left(\frac{1^{3}}{3}-1^{-1}+1\right)$$

Whakarūnātia.

$$\frac{34}{3}$$