№1
[tex]\displaystyle \int\limits x \cdot (2x-3)^8 \, dx= \frac{1}{4} \bigg ( \frac{(2x-3)^{10}}{10}+ \frac{(2x-3)^{9}}{3} \bigg )+C[/tex]
Введем замену
[tex]t = 2x - 3 \\\\ \dfrac{t}{2} = x -\dfrac{3}{2} \Leftrightarrow x= \dfrac{t+3}{2}[/tex]
[tex]\boxed{x\cdot (2x-3)^8 =\dfrac{1}{2} (t+3)\cdot t^8 = \dfrac{1}{2}(t^9 + 3 t^8)}[/tex]
[tex]\displaystyle \int\limits x \cdot (2x-3)^8 \, dx = \int\limits x \cdot (2x-3)^8 \cdot \frac{1}{2} \, d(2x-3) = \displaystyle \int\limits \frac{1}{2}\cdot (t^9 +3t^8) \cdot \frac{1}{2} \, dt = \\\\\\ =\frac{1}{4}\cdot \int\limits ( t^9 +3t^8)\, dt = \frac{1}{4}\bigg (\frac{t^{9+1}}{9+1}+ 3\cdot \frac{t^{8+1}}{8+1} \bigg) = \frac{1}{4}\bigg ( \frac{t^{10}}{10}+ \frac{t^9}{3 } \bigg )+C[/tex]
Подставим [tex]t = 2x - 3[/tex]
[tex]\displaystyle \frac{1}{4}\bigg ( \frac{t^{10}}{10}+ \frac{t^9}{3 } \bigg )+C = \frac{1}{4} \bigg ( \frac{(2x-3)^{10}}{10}+ \frac{(2x-3)^{9}}{3} \bigg )+C[/tex]
№2
[tex]\displaystyle \int\limits x \cdot (1-2x)^5 \, dx=\dfrac{1}{4} \bigg (\frac{(1-2x)^7}{7}-\frac{(1-2x)^6}{6} \bigg ) +C[/tex]
По той же аналогии
[tex]t = 1 - 2x \\\\ \dfrac{t}{2} = -x +\dfrac{1}{2} \Leftrightarrow x= \dfrac{1-t}{2}[/tex]
[tex]\boxed{\displaystyle x \cdot (1-2x)^5 =\frac{1-t}{2}\cdot t^5 = \frac{t^5 -t^6}{2}}[/tex]
[tex]\displaystyle \int\limits x \cdot (1-2x)^5 \, dx = \int\limits x \cdot (1-2x)^5 \cdot \bigg(-\frac{1}{2} \bigg )\, d(1-2x) =\\\\\\= \displaystyle \int\limits \frac{1}{2}\cdot\bigg(-\frac{1}{2} \bigg )\cdot (t^5 -t^6) \, dt = - \frac{1}{4}\int\limits (t^5 - t^6) \, dt = -\frac{1}{4}\bigg ( \frac{t^6}{6} - \frac{t^7}{7} \bigg) +C[/tex]
Подставим [tex]t = 1 - 2x[/tex]
[tex]\displaystyle -\frac{1}{4}\bigg ( \frac{t^6}{6} - \frac{t^7}{7} \bigg) +C = -\frac{1}{4}\bigg (\frac{(1-2x)^6}{6} -\frac{(1-2x)^7}{7} \bigg ) +C =\\\\\\ =\dfrac{1}{4} \bigg (\frac{(1-2x)^7}{7}-\frac{(1-2x)^6}{6} \bigg ) +C[/tex]
