Question

Найти неопределённые интегралы методом замены переменной

Answer 1

$t= \sqrt{3x+5}\\ \\ dt=\frac{3}{2\sqrt{3x+5}} \, dx \\ \\ t^2 =3x+5 \\ \\ x=\frac{1}{3}\cdot (t^2-5)$

$\int {((\frac{1}{3}\cdot (t^2-5))^2\cdot \frac{2}{3}t^2)} \, dt=\frac{2}{27} \int {((t^4-10t^2+25)\cdot t^2)} \, dt=\frac{2}{27}\int {(t^6-10t^4+25t^2)} \, dt= \\ \\ =\frac{2}{27}\cdot (\frac{t^7}{7}-\frac{10t^5}{5}+\frac{25t^3}{3})+C=\frac{2}{27}\cdot (\frac{t^7}{7}-2t^5+\frac{25t^3}{3})+C=\\\\=\frac{2}{27}\cdot\frac{1}{21}\cdot t^3 \cdot (3t^4-42t^2+175) =\frac{2}{567}\cdot t^3\cdot (3t^4-42t^2+175) \\\\ = \frac{2}{567}\cdot (\sqrt{3x+5})^3\cdot (3\cdot (\sqrt{3x+5})^4-42\cdot (\sqrt{3x+5})^2+175)$

$=\frac{2}{567}\cdot \sqrt{(3x+5)^3}\cdot (3\cdot (3x+5)^2-42\cdot (3x+5)+175)= \\ \\ =\frac{2}{567}\cdot \sqrt{(3x+5)^3}\cdot (3\cdot (9x^2+30x+25)-126x-210+175)= \\ \\ = \frac{2}{567} \cdot \sqrt{(3x+5)^3}\cdot (27x^2+90x+75-126x-35)=\\ \\ = \frac{2}{567} \cdot \sqrt{(3x+5)^3}\cdot (27x^2-36x+40)$