Whakaoti i te ∫ 3y/1+j^2 wrt y

$$\int \frac { 3 y } { 1 + j ^ { 2 } } d y$$

Aromātai

$$\frac{3y^{2}}{2\left(j^{2}+1\right)}+С$$

Ngā Upane Otinga

Whakatauwehetia te pūmau mā te whakamahi i te \int af\left(y\right)\mathrm{d}y=a\int f\left(y\right)\mathrm{d}y.

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

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

$$\frac{3}{1+j^{2}}\times \frac{y^{2}}{2}$$

Whakarūnātia.

$$\frac{3y^{2}}{2\left(1+j^{2}\right)}$$

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

$$\frac{3y^{2}}{2\left(1+j^{2}\right)}+С$$

Kimi Pārōnaki e ai ki y

$$\frac{3y}{j^{2}+1}$$