Whakaoti i te ∫ (2t+5x)(2t-5x)

$$\int ( 2 t + 5 x ) ( 2 t - 5 x )$$

Aromātai

$$4xt^{2}-\frac{25x^{3}}{3}+С$$

Ngā Upane Otinga

Whakaarohia te \left(2t+5x\right)\left(2t-5x\right). Ka taea te whakareanga te panoni ki te rerekētanga o ngā pūrua mā te ture: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\int \left(2t\right)^{2}-\left(5x\right)^{2}\mathrm{d}x$$

Whakarohaina te \left(2t\right)^{2}.

$$\int 2^{2}t^{2}-\left(5x\right)^{2}\mathrm{d}x$$

Tātaihia te 2 mā te pū o 2, kia riro ko 4.

$$\int 4t^{2}-\left(5x\right)^{2}\mathrm{d}x$$

Whakarohaina te \left(5x\right)^{2}.

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

Tātaihia te 5 mā te pū o 2, kia riro ko 25.

$$\int 4t^{2}-25x^{2}\mathrm{d}x$$

Kōmitimititia te kīanga tapeke mā te kīanga.

$$\int 4t^{2}\mathrm{d}x+\int -25x^{2}\mathrm{d}x$$

Whakatauwehea te pūmau i ēnei kīanga katoa.

$$4\int t^{2}\mathrm{d}x-25\int x^{2}\mathrm{d}x$$

Kimihia te tau tōpū o t^{2} mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}x=ax.

$$4t^{2}x-25\int x^{2}\mathrm{d}x$$

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}. Whakareatia -25 ki te \frac{x^{3}}{3}.

$$4t^{2}x-\frac{25x^{3}}{3}$$

Mēnā ko F\left(x\right) he pārōnaki kōaro o f\left(x\right), kāti ko te huinga o ngā pārōnaki kōaro katoa o f\left(x\right) ka whakaaturia e F\left(x\right)+C. Nō reira, me tāpiri te pūmau o te whakatōpūtanga C\in \mathrm{R} ki te otinga.

$$4t^{2}x-\frac{25x^{3}}{3}+С$$

Kimi Pārōnaki e ai ki x

$$4t^{2}-25x^{2}$$