∫ (4x^3-1/x^2) wrt x=x^4+1/x+C eching

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

C uchun yechish

$$C=С$$

$$x\neq 0$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini x ga ko'paytirish.

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

Ayni asosning daraja ko‘rsatkichlarini ko‘paytirish uchun ularning darajalarini qo‘shing. 1 va 4 ni qo‘shib, 5 ni oling.

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

Ifodalarni qo‘shish yoki ayirish uchun ularni yoyib, maxrajlarini bir xil qiling. 4x^{3} ni \frac{x^{2}}{x^{2}} marotabaga ko'paytirish.

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

\frac{4x^{3}x^{2}}{x^{2}} va \frac{1}{x^{2}} da bir xil maxraji bor, ularning suratini ayirish orqali ayiring.

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

4x^{3}x^{2}-1 ichidagi ko‘paytirishlarni bajaring.

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

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

Ikkala tarafdan x^{5} ni ayirish.

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

Ikkala tarafdan 1 ni ayirish.

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

Tenglama standart shaklda.

$$xC=Сx$$

Ikki tarafini x ga bo‘ling.

$$\frac{xC}{x}=\frac{Сx}{x}$$

x ga bo'lish x ga ko'paytirishni bekor qiladi.

$$C=\frac{Сx}{x}$$

Сx ni x ga bo'lish.

$$C=С$$

x uchun yechish

$$x\neq 0$$

$$C=С\text{ and }x\neq 0$$