Задача 37255 ...

yelymcheav

2019-05-19 12:10:04

Найти интеграл ∫ dx/(sqrt(x)(корень 3ей степени (x)+1))

sova

2019-05-19 12:17:40

Замена
x=t^6
dx=6t^5dt
sqrt(x)=t^3
∛x=t^2

= ∫ 6t^5dt/(t^3+t^2)=6* ∫ t^3dt/(t+1)=6* ∫ (t^3-1+1)dt/(t+1)=

=6* ∫ (t^3-1)dt/(t+1)+ 6* ∫ dt/(t+1)=

=6* ∫(t^2+t+1)dt + 6* ∫ dt/(t+1)=

=6*(t^3/3)+6*(t^2/2)+6t + 6ln|t+1|+C=

[b]=2t^3+3t^2+6t+6ln|t+1)+C, t=x^(1/6)[/b]


О т в е т. 2sqrt(x)+3∛x+6x^(1/6)+6ln|x^(1/6)+1|+C