Question

Помогите пожалуйста
​

Answer 1

Ответ:

$\int {tg}^{4} (x + 3)dx$

$\int {tg}^{4} (x + 3) \times \frac{1}{1} dt$

$\int {tg}^{4} (x + 3) \times 1dt$

$\int {tg}^{4} (x + 3)dt$

$\int {tg}^{4} (t)dt$

$\frac{1}{4 - 1} \times {tg}^{4 - 1} (t) - \int {tg}^{4 - 2} (t)dt$

$\frac{1}{3} \times {tg}^{3} (t) - \int {tg}^{2} (t)dt$

$\frac{1}{3} \times {tg}^{3} (t) - \int {sec}^{2} (t) - 1dt$

$\frac{1}{3} \times {tg}^{3} (t) - (\int {sec}^{2} (t)dt - \int1dt)$

$\frac{1}{3} \times {tg}^{3} (t) - (tg(t) - t)$

$\frac{1}{3} \times {tg}^{3} (x + 3) - (tg(x + 3) - (x + 3))$

$\frac{ {tg}^{3}(x + 3) }{3} - (tg(x + 3) - x - 3)$

$\frac{ {tg}^{3}(x + 3) }{3} - tg(x + 3) + x + 3$

$\frac{ {tg}^{3} (x + 3)}{3} - tg(x + 3) + x + 3 + C, C\in\R$

$\frac{ {tg}^{3} (x + 3)}{3} - tg(x + 3) + x + C, C\in\R$