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gubaevamilyaus
@gubaevamilyaus
July 2022
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Помогите решить задачу по интеграллам, пож.)
(x+1) dx / (2x+1)^1/3
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mikael2
∫(х+1)dx/(2x+1)^1/3
2x+1=z 2x=z-1 x=z/2-1/2 x+1=z/2+1/2 2dx=dz dx=dz/2
∫(z/2)*0.5dz/z^1/3+1/2∫0.5dz/z^1/3=
=1/4∫z^2/3 dz +1/4∫z^(-1/3)dz=
3/20*z^(5/3)+3/8z^(2/3)
+C=
=3/20*(2x+1)^(5/3)+3/8*(2x+1)^(2/3)+C
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Answers & Comments
2x+1=z 2x=z-1 x=z/2-1/2 x+1=z/2+1/2 2dx=dz dx=dz/2
∫(z/2)*0.5dz/z^1/3+1/2∫0.5dz/z^1/3=
=1/4∫z^2/3 dz +1/4∫z^(-1/3)dz=3/20*z^(5/3)+3/8z^(2/3)+C=
=3/20*(2x+1)^(5/3)+3/8*(2x+1)^(2/3)+C