Whakaoti i te ∫ (4x^3-1/x^2) wrt x=x^4+1/x+C

$$\int ( 4 x ^ { 3 } - \frac { 1 } { x ^ { 2 } } ) d x = x ^ { 4 } + \frac { 1 } { x } + C$$

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.

$$x\int 4x^{3}-\frac{1}{x^{2}}\mathrm{d}x=xx^{4}+1+xC$$

Hei whakarea i ngā pū o te pūtake kotahi, me tāpiri ō rātou taupū. Tāpiria te 1 me te 4 kia riro ai te 5.

$$x\int 4x^{3}-\frac{1}{x^{2}}\mathrm{d}x=x^{5}+1+xC$$

Hei tāpiri, hei tango kīanga rānei, me whakaroha ērā kia rite ā rātou tauraro. Whakareatia 4x^{3} ki te \frac{x^{2}}{x^{2}}.

$$x\int \frac{4x^{3}x^{2}}{x^{2}}-\frac{1}{x^{2}}\mathrm{d}x=x^{5}+1+xC$$

Tā te mea he rite te tauraro o \frac{4x^{3}x^{2}}{x^{2}} me \frac{1}{x^{2}}, me tango rāua mā te tango i ō raua taurunga.

$$x\int \frac{4x^{3}x^{2}-1}{x^{2}}\mathrm{d}x=x^{5}+1+xC$$

Mahia ngā whakarea i roto o 4x^{3}x^{2}-1.

$$x\int \frac{4x^{5}-1}{x^{2}}\mathrm{d}x=x^{5}+1+xC$$

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

$$x^{5}+1+xC=x\int \frac{4x^{5}-1}{x^{2}}\mathrm{d}x$$

Tangohia te x^{5} mai i ngā taha e rua.

$$1+xC=x\int \frac{4x^{5}-1}{x^{2}}\mathrm{d}x-x^{5}$$

Tangohia te 1 mai i ngā taha e rua.

$$xC=x\int \frac{4x^{5}-1}{x^{2}}\mathrm{d}x-x^{5}-1$$

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

$$x\neq 0$$

$$C=С\text{ and }x\neq 0$$