Задача 51854 ...

sardor51819

2020-05-28 16:43:57

[m]\int_{0}^{ ∞ }[/m]xdx/[m]\sqrt[4]{(16+x^2)^5}[/m]

sova

2020-05-28 17:02:36

[m] ∫ u^{-\frac{5}{4}}du=\frac{u^{-\frac{5}{4}+1}}{-\frac{5}{4}+1}[/m]


u=16+x^2

du=(16+x^2)`*dx=2xdx ⇒

xdx=(1/2)du


[b]xdx=(1/2)d(16+x^2)[/b]


=[m] \frac{1}{2}[/m]∫ ^(+ ∞ )_(0)[m](16+x^2)^{-\frac{5}{4}}d(16+x^2)[/m]=

=[m]\frac{1}{2}\cdot \frac{u^{-\frac{5}{4}+1}}{-\frac{5}{4}+1}[/m]|^(+ ∞ )_(0)=


=[m]-\frac{2}{\sqrt[4]{16+x^2}}[/m]|^(+ ∞ )_(0)=-0+1=1