Задача 71866 Нужно решить ...
Условие
Решение
[m]\sqrt[4]{5x+1}=t[/m] ⇒ [m]5x+1=t^4[/m] ⇒ [m]5x=t^4-1[/m] ⇒ [m]x=\frac{t^4-1}{5}[/m]
[m]dx=\frac{4}{5}t^3dt[/m]
x=0
[m]\sqrt[4]{5\cdot 0+1}=t[/m] ⇒t=1
x=3
[m]\sqrt[4]{5\cdot 3+1}=t[/m] ⇒t=2
[m] ∫^{3} _{0}\frac{15xdx}{\sqrt[4]{(5x+1)^3}+\sqrt[4]{5x+1}}= ∫ ^{2}_{1}\frac{15\cdot \frac{t^4-1}{5}\cdot \frac{4}{5}t^3dt}{t^3+t}=\frac{12}{5}∫ ^{2}_{1}\frac{(t^4-5)\cdot t^3}{t^3+t}dt=\frac{12}{5}∫ ^{2}_{1}\frac{(t^4-5)\cdot t^3}{t(t^2+1)}dt=\frac{12}{5}∫ ^{2}_{1}\frac{(t^4-5)\cdot t^2}{t^2+1}dt=\frac{12}{5}∫ ^{2}_{1}\frac{t^6-5\cdot t^2}{t^2+1}dt=[/m]
(см. интегрирование рациональных дробей)
[m]=\frac{12}{5}\cdot (4 arctg t +\frac{t^5}{5}-\frac{t^3}{3}-4t)|^{2}_{1}=\frac{12}{5}\cdot (4 arctg 2 +\frac{2^5}{5}-\frac{2^3}{3}-8)-\frac{12}{5}\cdot (4 arctg 1 +\frac{1^5}{5}-\frac{1^3}{3}-4)=\frac{12}{5}\cdot (4arctg2-π-\frac{2}{15})[/m]
