Whakaoti i te ∫ wrt x/x^2=-1/x+c

$$\int \frac { d x } { x ^ { 2 } } = - \frac { 1 } { x } + c$$

Whakaoti mō c

$$c=С$$

$$x\neq 0$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Whakareatia ngā taha e rua o te whārite ki te x.

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

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

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

Me tāpiri te 1 ki ngā taha e rua.

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

He hanga arowhānui tō te whārite.

$$xc=Сx$$

Whakawehea ngā taha e rua ki te x.

$$\frac{xc}{x}=\frac{Сx}{x}$$

Mā te whakawehe ki te x ka wetekia te whakareanga ki te x.

$$c=\frac{Сx}{x}$$

Whakawehe Сx ki te x.

$$c=С$$

Whakaoti mō x

$$x\neq 0$$

$$c=С\text{ and }x\neq 0$$