Задача 22931 ...

u10919724856

27.01.2018

∫ (5-∛(х^2 )) / √х dx

vk397114329

27.01.2018

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Решение: ∫ (5-∛x^2)/ sqrt(x)dx=
= ∫ 5dx/ sqrt(x)- ∫ x^1/6dx=10xsgrt(x)/3-6x^7/6/7+c

SOVA

28.01.2018

∛(х^2)=x^(2/3)
sqrt(x)=x^(1/2)
1/(x^(1/2))=x^(-1/2)

(5-∛(х^2))/sqrt(x)=(5-x^(2/3))/x^(1/2)=(5-x^(2/3))*x^(-1/2)=
=5x^(-1/2)-x^((2/3)*x^((-1/2))=
=5x^(-1/2)-x^((4/6)-(3/6))=
=5x^(-1/2)-x^(1/6)


∫ (5-∛(х^2))dx/sqrt(x) =∫ (5x^(-1/2)-x^(1/6))dx=

=5*x^((-1/2)+1)/((-1/2)+1) - (x^(1/6)+1)/((1/6)+1) + C=

=5*x^(1/2)/(1/2)) - (x^(7/6))/(7/6) + C=

=10sqrt(x) -(6/7)*x*x^(1/6) + C