u=tg5x–2
[m]d(tg5x-2)=(tg5x-2)`dx=\frac{1}{cos^25x}\cdot (5x)`dx=\frac{5}{cos^25x}dx[/m]
[m] ∫ \frac{\sqrt[3]{(tg5x-2)^2}}{cos^25x}dx=\frac{1}{5}∫ \frac{5\sqrt[3]{(tg5x-2)^2}}{cos^25x}dx=frac{1}{5}∫(tg5x-2)^2d(tg5x-2)=\frac{1}{5}\cdot \frac{(tg5x-2)^{\frac{2}{3}+1}}{\frac{2}{3}+1}+C=[/m] можно упростить
Интегрирование по частям
[m]u=(3x+4)[/m] ⇒ [m]du=3dx[/m]
[m]dv=9^{x}dx[/m] ⇒ [m]v] ∫ dv= ∫ 9^{x}dx=\frac{9^{x}}{ln9}[/m]
[m] ∫ (3x+4)9^{x}dx=(3x+4)\cdot\frac{9^{x}}{ln9} - ∫\frac{9^{x}}{ln9} 3 dx=(3x+4)\cdot\frac{9^{x}}{ln9}-\frac{3}{ln9}\cdot \frac{9^{x}}{ln9}+C[/m] можно упростить